Placeholder Image

字幕列表 影片播放

  • VOICEOVER: The following content is provided under a Creative

  • Commons license.

  • Your support will help MIT Open Courseware

  • continue to offer high-quality educational resources for free.

  • To make a donation or to view additional materials

  • from hundreds of MIT courses, visit MIT OpenCourseWare

  • at ocw.mit.edu.

  • JULIAN SHUN: Good afternoon, everyone.

  • So today we're going to talk about storage allocation.

  • This is a continuation from last lecture

  • where we talked about serial storage allocation.

  • Today we'll also talk a little bit more

  • about serial allocation.

  • But then I'll talk more about parallel allocation and also

  • garbage collection.

  • So I want to just do a review of some memory allocation

  • primitives.

  • So recall that you can use malloc to allocate memory

  • from the heap.

  • And if you call malloc with the size of s,

  • it's going to allocate and return

  • a pointer to a block of memory containing at least s bytes.

  • So you might actually get more than s bytes,

  • even though you asked for s bytes.

  • But it's guaranteed to give you at least s bytes.

  • The return values avoid star, but good programming practice

  • is to typecast this pointer to whatever type

  • you're using this memory for when you receive

  • this from the malloc call.

  • There's also aligned allocation.

  • So you can do aligned allocation with memalign,

  • which takes two arguments, a size a as well as a size s.

  • And a has to be an exact power of 2,

  • and it's going to allocate and return

  • a pointer to a block of memory again containing

  • at least s bytes.

  • But this time this memory is going

  • to be aligned to a multiple of a,

  • so the address is going to be a multiple of a,

  • where this memory block starts.

  • So does anyone know why we might want to do an aligned memory

  • allocation?

  • Yeah?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yeah, so one reason is

  • that you can align memories so that they're

  • aligned to cache lines, so that when you access an object that

  • fits within the cache line, it's not

  • going to cross two cache lines.

  • And you'll only get one cache axis instead of two.

  • So one reason is that you want to align

  • the memory to cache lines to reduce

  • the number of cache misses.

  • You get another reason is that the vectorization operators

  • also require you to have memory addresses that

  • are aligned to some power of 2.

  • So if you align your memory allocation with memalign,

  • then that's also good for the vector units.

  • We also talked about deallocations.

  • You can free memory back to the heap with the free function.

  • So if you pass at a point of p to some block of memory,

  • it's going to deallocate this block

  • and return it to the storage allocator.

  • And we also talked about some anomalies of freeing.

  • So what is it called when you fail to free

  • some memory that you allocated?

  • Yes?

  • Yeah, so If you fail to freeze something that you allocated,

  • that's called a memory leak.

  • And this can cause your program to use more and more memory.

  • And eventually your program is going

  • to use up all the memory on your machine,

  • and it's going to crash.

  • We also talked about freeing something more than once.

  • Does anyone remember what that's called?

  • Yeah?

  • Yeah, so that's called double freeing.

  • Double freeing is when you free something more than once.

  • And the behavior is going to be undefined.

  • You might get a seg fault immediately,

  • or you'll free something that was allocated

  • for some other purpose.

  • And then later down the road your program

  • is going to have some unexpected behavior.

  • OK.

  • I also want to talk about m map.

  • So m map is a system call.

  • And usually m map is used to treat some file on disk

  • as part of memory, so that when you

  • write to that memory region, it also backs it up on disk.

  • In this context here, I'm actually

  • using m map to allocate virtual memory without having

  • any backing file.

  • So

  • So our map has a whole bunch of parameters here.

  • The second to the last parameter indicates

  • the file I want to map, and if I pass a negative 1,

  • that means there's no backing file.

  • So I'm just using this to allocate some virtual memory.

  • The first argument is where I want to allocate it.

  • And 0 means that I don't care.

  • The size in terms of number of bytes

  • has how much memory I want to allocate.

  • Then there's also permissions.

  • So here it says I can read and write this memory region.

  • s private means that this memory region

  • is private to the process that's allocating it.

  • And then map anon means that there is no name associated

  • with this memory region.

  • And then as I said, negative 1 means

  • that there's no backing file.

  • And the last parameter is just 0 if there's no backing file.

  • Normally it would be an offset into the file

  • that you're trying to map.

  • But here there's no backing file.

  • And what m map does is it finds a contiguous unused region

  • in the address space of the application that's large enough

  • to hold size bytes.

  • And then it updates the page table

  • so that it now contains an entry for the pages

  • that you allocated.

  • And then it creates a necessary virtual memory management

  • structures within the operating system

  • to make it so that users accesses to this area

  • are legal, and accesses won't result in a seg fault.

  • If you try to access some region of memory without using--

  • without having OS set these parameters,

  • then you might get a set fault because the program might not

  • have permission to access that area.

  • But m map is going to make sure that the user can access

  • this area of virtual memory.

  • And m map is a system call, whereas malloc,

  • which we talked about last time, is a library call.

  • So these are two different things.

  • And malloc actually uses m map under the hood

  • to get more memory from the operating system.

  • So let's look at some properties of m map.

  • So m map is lazy.

  • So when you request a certain amount of memory,

  • it doesn't immediately allocate physical memory

  • for the requested allocation.

  • Instead it just populates the page table

  • with entries pointing to a special 0 page.

  • And then it marks these pages as read only.

  • And then the first time you write to such a page,

  • it will cause a page fault. And at that point,

  • the OS is going to modify the page table,

  • get the appropriate physical memory,

  • and store the mapping from the virtual address space

  • to physical address space for the particular page

  • that you touch.

  • And then it will restart the instructions

  • so that it can continue to execute.

  • You can-- turns out that you can actually

  • m map a terabyte of virtual memory,

  • even on a machine with just a gigabyte of d ram.

  • Because when you call m map, it doesn't actually

  • allocate the physical memory.

  • But then you should be careful, because a process might

  • die from running out of physical memory

  • well after you call m map.

  • Because m map is going to allocate this physical memory

  • whenever you first touch it.

  • And this could be much later than when you actually

  • made the call to m map.

  • So any questions so far?

  • OK.

  • So what's the difference between malloc and m map?

  • So as I said, malloc is a library call.

  • And it's part of--malloc and free are part of the memory

  • allocation interface of the heat-management code in the c

  • library.

  • And the heat-management code uses the available system

  • facilities, including the m map function

  • to get a virtual address space from the operating system.

  • And then the heat-management code

  • is going-- within malloc-- is going

  • to attempt to satisfy user requests for heat storage

  • by reusing the memory that it got from the OS

  • as much as possible until it can't do that anymore.

  • And then it will go and call m map

  • to get more memory from the operating system.

  • So the malloc implementation invokes m map and other system

  • calls to expand the size of the users heap storage.

  • And the responsibility of malloc is

  • to reuse the memory, such that your fragmentation is reduced,

  • and you have good temporal locality,

  • whereas the responsibility of m map

  • is actually getting this memory from the operating system.

  • So any questions on the differences

  • between malloc and m map?

  • So one question is, why don't we just call

  • m map up all the time, instead of just using malloc?

  • Why don't we just directly call m map?

  • Yes.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yes, so one answer is

  • that you might have free storage from before

  • that you would want to reuse.

  • And it turns out that m map is relatively heavy weight.

  • So it works on a page granularity.

  • So if you want to do a small allocation,

  • it's quite wasteful to allocate an entire page

  • for that allocation and not reuse it.

  • You'll get very bad external fragmentation.

  • And when you call m map, it has to go

  • through all of the overhead of the security of the OS

  • and updating the page table and so on.

  • Whereas, if you use malloc, it's actually

  • pretty fast for most allocations,

  • and especially if you have temporal locality where

  • you allocate something that you just recently freed.

  • So your program would be pretty slow

  • if you used m map all the time, even for small allocations.

  • For big allocations, it's fine.

  • But for small allocations, you should use malloc.

  • Any questions on m map versus malloc?

  • OK, so I just want to do a little bit of review

  • on how address translation works.

  • So some of you might have seen this before in your computer

  • architecture course.

  • So how it works is, when you access memory location,

  • you access it via the virtual address.

  • And the virtual address can be divided into two parts, where

  • the lower order bits store the offset,

  • and the higher order bits store the virtual page number.

  • And in order to get the physical address associated

  • with this virtual address, the hardware

  • is going to look up this virtual page number in what's

  • called the page table.

  • And then if it finds a corresponding entry

  • for the virtual page number in the page table,

  • that will tell us the physical frame number.

  • And then the physical frame number

  • corresponds to where this fiscal memory is in d ram.

  • So you can just take the frame number,

  • and then use the same offset as before

  • to get the appropriate offset into the physical memory frame.

  • So if the virtual page that you're looking for

  • doesn't reside in physical memory,

  • then a page fault is going to occur.

  • And when a page fault occurs, either the operating system

  • will see that the process actually

  • has permissions to look at that memory region,

  • and it will set the permissions and place the entry

  • into the page table so that you can

  • get the appropriate physical address.

  • But otherwise, the operating system

  • might see that this process actually

  • can't access that region memory, and then you'll

  • get a segmentation fault.

  • It turns out that the page table search, also called a page

  • walk, is pretty expensive.

  • And that's why we have the translation look, a side

  • buffer or TLB, which is essentially

  • a cache for the page table.

  • So the hardware uses a TLB to cache the recent page

  • table look ups into this TLB so that later on when

  • you access the same page, it doesn't

  • have to go all the way to the page table

  • to find the physical address.

  • It can first look in the TLB to see

  • if it's been recently accessed.

  • So why would you expect to see something

  • that it recently has accessed?

  • So what's one property of a program

  • that will make it so that you get a lot of TLB hits?

  • Yes?

  • STUDENT: Well, usually [INAUDIBLE] nearby one another,

  • which means they're probably in the same page or [INAUDIBLE]..

  • JULIAN SHUN: Yeah, so that's correct.

  • So the page table stores pages, which

  • are typically four kilobytes.

  • Nowadays there are also huge pages, which

  • can be a couple of megabytes.

  • And most of the accesses in your program

  • are going to be near each other.

  • So they're likely going to reside

  • on the same page for accesses that have been

  • done close together in time.

  • So therefore you'll expect that many of your recent accesses

  • are going to be stored in the TLB

  • if your program has locality, either spatial or temporal

  • locality or both.

  • So how this architecture works is that the processor is first

  • going to check whether the virtual address you're

  • looking for is in TLB.

  • If it's not, it's going to go to the page table and look it up.

  • And then if it finds that there, then it's

  • going to store that entry into the TLB.

  • And then next it's going to go get this physical address

  • that it found from the TLB and look it up into the CPU cache.

  • And if it finds it there, it gets it.

  • If it doesn't, then it goes to d ram to satisfy the request.

  • Most modern machines actually have an optimization

  • that allow you to do TLB access in parallel with the L1 cache

  • access.

  • So the L1 cache actually uses virtual addresses instead

  • of fiscal addresses, and this reduces

  • the latency of a memory access.

  • So that's a brief review of address translation.

  • All right, so let's talk about stacks.

  • So when you execute a serial c and c++ program,

  • you're using a stack to keep track of the function calls

  • and local variables that you have to save.

  • So here, let's say we have this invocation tree,

  • where function a calls Function b, which then returns.

  • And then a calls function c, which

  • calls d, returns, calls e, returns, and then returns

  • again.

  • Here are the different views of the stack at different points

  • of the execution.

  • So initially when we call a, we have a stack frame for a.

  • And then when a calls b, we're going

  • to place a stack frame for b right

  • below the stack frame of a.

  • So these are going to be linearly ordered.

  • When we're done with b, then this part of the stack

  • is no longer going to be used, the part for b.

  • And then when it calls c, It's going to allocate a stack frame

  • below a on the stack.

  • And this space is actually going to be the same space as what

  • b was using before.

  • But this is fine, because we're already done with the call

  • to b.

  • Then when c calls d, we're going to create a stack frame for d

  • right below c.

  • When it returns, we're not going to use that space any more,

  • so then we can reuse it for the stack frame when we call e.

  • And then eventually all of these will pop back up.

  • And all of these views here share the same view

  • of the stack frame for a.

  • And then for c, d, and e, they all stare share the same view

  • of this stack for c.

  • So this is how a traditional linear stack works when you

  • call a serial c or c++ program.

  • And you can view this as a serial walk over the invocation

  • tree.

  • There's one rule for pointers.

  • With traditional linear stacks is

  • that a parent can pass pointers to its stack variables

  • down to its children.

  • But not the other way around.

  • A child can't pass a pointer to some local variable

  • back to its parent.

  • So if you do that, you'll get a bug in your program.

  • How many of you have tried doing that before?

  • Yeah, so a lot of you.

  • So let's see why that causes a problem.

  • So if I'm calling--

  • if I call b, and I pass a pointer to some local variable

  • in b stack to a, and then now when a calls c,

  • It's going to overwrite the space that b was using.

  • And if b's local variable was stored in the space

  • that c has now overwritten, then you're

  • just going to see garbage.

  • And when you try to access that, you're

  • not going to get the correct value.

  • So you can pass a pointer to a's local variable

  • down to any of these descendant function

  • calls, because they all see the same view of a stack.

  • And that's not going to be overwritten

  • while these descendant function calls are proceeding.

  • But if you pass it the other way, then potentially

  • the variable that you had a pointer to

  • is going to be overwritten.

  • So here's one question.

  • If you want to pass memory from a child back to the parent,

  • where would you allocate it?

  • So you can allocate it on the parent.

  • What's another option?

  • Yes?

  • Yes, so another way to do this is to allocate it on the heap.

  • If you allocate it on the heap, even after you

  • return from the function call, that memory

  • is going to persist.

  • You can also allocate it in the parent's stack, if you want.

  • In fact, some programs are written that way.

  • And one of the reasons why many c functions require

  • you to pass in memory to the function where it's

  • going to store the return value is

  • to try to avoid an expensive heap allocation in the child.

  • Because if the parent allocates this space to store the result,

  • the child can just put whatever it

  • wants to compute in that space.

  • And the parent will see it.

  • So then the responsibility is up to the parent

  • to figure out whether it wants to allocate the memory

  • on the stack or on the heap.

  • So this is one of the reasons why

  • you'll see many c functions, where one of the arguments

  • is a memory location where the result should be stored.

  • OK, so that was the serial case.

  • What happens in parallel?

  • So in parallel, we have what's called

  • a cactus stack where we can support multiple views

  • of the stack in parallel.

  • So let's say we have a program where it calls function

  • a, and then a spawns b and c.

  • So b and c are going to be running potentially

  • in parallel.

  • And then c spawns d and e, which can potentially

  • be running in parallel.

  • So for this program, we could have functions b, d and e all

  • executing in parallel.

  • And a cactus stack is going to allow

  • us to have all of these functions

  • see the same view of this stack as they

  • would have if this program were executed serially.

  • And the silk runtime system supports

  • the cactus stack to make it easy for writing parallel programs.

  • Because now when you're writing programs,

  • you just have to obey the same rules for programming in serial

  • c and c++ with regards to the stack,

  • and then you'll still get the intended behavior.

  • And it turns out that there's no copying of the stacks here.

  • So all of these different views are

  • seeing the same virtual memory addresses for a.

  • But now there is an issue of how do

  • we implement this cactus stack?

  • Because in the serial case, we could have these later stacks

  • overwriting the earlier stacks.

  • But in parallel, how can we do this?

  • So does anyone have any simple ideas

  • on how we can implement a cactus stack?

  • Yes?

  • STUDENT: You could just have each child's stack start

  • in like a separate stack, or just have references

  • to the [INAUDIBLE].

  • JULIAN SHUN: Yeah, so one way to do this

  • is to have each thread use a different stack.

  • And then store pointers to the different stack frames

  • across the different stacks.

  • There's actually another way to do this, which is easier.

  • OK, yes?

  • STUDENT: If the stack frames have a maximum--

  • fixed maximum size-- then you could put them

  • all in the same stack separated by that fixed size.

  • JULIAN SHUN: Yeah, so if the stacks all

  • have a maximum depth, then you could just

  • allocate a whole bunch of stacks, which are separated

  • by this maximum depth.

  • There's actually another way to do this,

  • which is to not use the stack.

  • So yes?

  • STUDENT: Could you memory map it somewhere else--

  • each of the different threads?

  • JULIAN SHUN: Yes, that's actually one way to do it.

  • The easiest way to do it is just to allocate it off the heap.

  • So instead of allocating the frames on the stack,

  • you just do a heap allocation for each of these stack frames.

  • And then each of these stack frames

  • has a pointer to the parent stack frame.

  • So whenever you do a function call,

  • you're going to do a memory allocation from the heap

  • to get a new stack frame.

  • And then when you finish a function,

  • you're going to pop something off of this stack,

  • and free it back to the heap.

  • In fact, a lot of early systems for parallel programming

  • use this strategy of heap-based cactus stacks.

  • Turns out that you can actually minimize the performance

  • impact using this strategy if you optimize the code enough.

  • But there is actually a bigger problem

  • with using a heap-based cactus stack, which doesn't

  • have to do with performance.

  • Does anybody have any guesses of what this potential issue is?

  • Yeah?

  • STUDENT: It requires you to allocate the heap in parallel.

  • JULIAN SHUN: Yeah, so let's assume that we can

  • do parallel heap allocation.

  • And we'll talk about that.

  • So assuming that we can do that correctly,

  • what's the issue with this approach?

  • Yeah?

  • STUDENT: It's that you don't know how big the stack is

  • going to be?

  • JULIAN SHUN: So let's assume that you

  • can get whatever stack frames you need from the heap,

  • so you don't actually need to put an upper bound on this.

  • Yeah?

  • STUDENT: We don't know the maximum depth.

  • JULIAN SHUN: Yeah.

  • So we don't know the maximum depth,

  • but let's say we can make that work.

  • So you don't actually need to know the maximum depth

  • if you're allocating off the heap.

  • Any other guesses?

  • Yeah?

  • STUDENT: Something to do with returning

  • from the stack that is allocated on the heap

  • to one of the original stacks.

  • JULIAN SHUN: So let's say we could get that to work as well.

  • So what happens if I try to run some program using

  • this heap-based cactus stack with something

  • that's using the regular stack?

  • So let's say I have some old legacy

  • code that was already compiled using

  • the traditional linear stack.

  • So there's a problem with interoperability here.

  • Because the traditional code is assuming

  • that, when you make a function call,

  • the stack frame for the function call

  • is going to appear right after the stack frame

  • for the particular call e function.

  • So if you try to mix code that uses the traditional stack as

  • well as this heap-based cactus stack approach,

  • then it's not going to work well together.

  • One approach is that you can just

  • recompile all your code to use this heap-based cactus stack.

  • Even if you could do that, even if all of the source codes

  • were available, there are some legacy programs

  • that actually in the source code,

  • they do some manipulations with the stack,

  • because they assume that you're using the traditional stack,

  • and those programs would no longer

  • work if you're using a heap-based cactus stack.

  • So the problem is interoperability

  • with legacy code.

  • Turns out that you can fix this using an approach

  • called thread local memory mapping.

  • So one of the students mentioned memory mapping.

  • But that requires changes to the operating system.

  • So it's not general purpose.

  • But the heap-based cactus stack turns out to be very simple.

  • And we can prove nice bounds about it.

  • So besides the interoperability issue,

  • heap-based cactus stacks are pretty good in practice,

  • as well as in theory.

  • So we can actually prove a space bound

  • of a cilk program that uses the heap-based cactus stack.

  • So let's say s 1 is the stack space required

  • by a serial execution of a cilk program.

  • Then the stack space of p worker execution

  • using a heap-based cactus stack is going to be

  • upper bounded by p times s 1.

  • So s p is the space for a p worker execution,

  • and that's less than or equal to p times s 1.

  • To understand how this works, we need

  • to understand a little bit about how the cilks works

  • stealing algorithm works.

  • So in the cilk work-stealing algorithm,

  • whenever you spawn something of work,

  • or that spawns a new task, is going to work on the task

  • that it spawned.

  • So therefore, for any leaf in the invocation tree that

  • currently exists, there's always going

  • to be a worker working on it.

  • There's not going to be any leaves in the tree

  • where there's no worker working on it.

  • Because when a worker spawns a task, it creates a new leaf.

  • But then it works immediately on that leaf.

  • So here we have a--

  • we have a invocation tree.

  • And for all of the leaves, we have a processor working on it.

  • And with this busy leaves property,

  • we can easily show this space bound.

  • So for each one of these processors,

  • the maximum stack space it's using

  • is going to be upper bounded by s 1,

  • because that's maximum stock space across a serial execution

  • that executes the whole program.

  • And then since we have p of these leaves,

  • we just multiply s 1 by p, and that gives us

  • an upper bound on the overall space used by a p worker

  • execution.

  • This can be a loose upper bound, because we're double

  • counting here.

  • There's some part of this memory that we're

  • counting more than once, because they're shared

  • among the different processors.

  • But that's why we have the less than or equal to here.

  • So it's upper bounded by p times s 1.

  • So this is one of the nice things

  • about using a heap-based cactus stack is that you

  • get this good space bound.

  • Any questions on the space bound here?

  • So let's try to apply this theorem to a real example.

  • So this is the divide and conquer matrix multiplication

  • code that we saw in a previous lecture.

  • So this is-- in this code, we're making eight recursive calls

  • to a divide and conquer function.

  • Each of size n over 2.

  • And before we make any of these calls,

  • we're doing a malloc to get some temporary space.

  • And this is of size order and squared.

  • And then we free this temporary space at the end.

  • And notice here that the allocations

  • of the temporary matrix obey a stack discipline.

  • So we're allocating stuff before we make recursive calls.

  • And we're freeing it after, or right before we

  • return from the function.

  • So all this stack--

  • all the allocations are nested, and they

  • follow a stack discipline.

  • And it turns out that even if you're allocating off

  • the heap, if you follow a stack discipline,

  • you can still use the space bound from the previous slide

  • to upper bound the p worker space.

  • OK, so let's try to analyze the space of this code here.

  • So first let's look at what the work and span are.

  • So this is just going to be review.

  • What's the work of this divide and conquer matrix multiply?

  • So it's n cubed.

  • So it's n cubed because we have eight solve problems

  • of size n over 2.

  • And then we have to do linear work

  • to add together the matrices.

  • So our recurrence is going to be t 1 of n

  • is equal to eight times t 1 of n over 2 plus order n squared.

  • And that solves to order n cubed if you just pull out

  • your master theorem card.

  • What about the span?

  • So what's the recurrence here?

  • Yeah, so the span t infinity of n

  • is equal to t infinitive of n over 2

  • plus a span of the addition.

  • And what's the span of the addition?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: No, let's assume that we

  • have a parallel addition.

  • We have nested silk four loops.

  • Right, so then the span of that is just going of be log n.

  • Since the span of 1 silk four loop is log n

  • and when you nest them, you just add together the span.

  • So it's going to be t infinity of n

  • is equal to t infinity of n over 2 plus order log n.

  • And what does that solve to?

  • Yeah, so it's going to solve to order log squared n.

  • Again you can pull out your master theorem card,

  • and look at one of the three cases.

  • OK, so now let's look at the space.

  • What's going to be the recurrence for the space?

  • Yes.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: The only place we're generating new space

  • is when we call this malloc here.

  • So they're all seeing the same original matrix.

  • So what would the recurrence be?

  • Yeah?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yeah.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: So the n square term is right.

  • Do we actually need eight subproblems of size n over 2?

  • What happens after we finish one of these sub problems?

  • Are we still going to use the space for it?

  • STUDENT: Yeah, you free the memory after the [INAUDIBLE]..

  • JULIAN SHUN: Right.

  • So you can actually reuse the memory.

  • Because you free the memory you allocated

  • after each one of these recursive calls.

  • So therefore the recurrence is just going to be s of n over 2

  • plus theta n squared.

  • And what does that solve to?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: N squared.

  • Right.

  • So here the n squared term actually dominates.

  • You have a decreasing geometric series.

  • So it's dominated at the root, and you get theta of n squared.

  • And therefore by using the busy leaves property and the theorem

  • for the space bound, this tells us that on p processors,

  • the space is going to be bounded by p times n squared.

  • And this is actually pretty good since we have a bound on this.

  • It turns out that we can actually prove a stronger bound

  • for this particular example.

  • And I'll walk you through how we can prove this stronger bound.

  • Here's the order p times n squared is already pretty good.

  • But we can actually do better if we look internally at how

  • this algorithm is structured.

  • So on each level of recursion, we're branching eight ways.

  • And most of the space is going to be

  • used near the top of this recursion tree.

  • So if I branch as much as possible

  • near the top of my recursion tree,

  • then that's going to give me my worst case space bound.

  • Because the space is decreasing geometrically as I

  • go down the tree.

  • So I'm going to branch eight ways

  • until I get to some level k in the recursion tree

  • where I have p nodes.

  • And at that point, I'm not going to branch anymore because I've

  • already used up all p nodes.

  • And that's the number of workers I have.

  • So let's say I have this level k here, where I have p nodes.

  • So what would be the value of k here?

  • If I branch eight ways how many levels do

  • I have to go until I get to p nodes?

  • Yes.

  • STUDENT: It's log base 8 of p.

  • JULIAN SHUN: Yes.

  • It's log base 8 of p.

  • So we have eight, the k, equal p,

  • because we're branching k ways.

  • And then using some algebra, you can get it

  • so that k is equal to log base 8 of p, which is equal to log

  • base 2 of p divided by 3.

  • And then at this level k downwards,

  • it's going to decrease geometrically.

  • So the space is going to be dominant at this level k.

  • So the space decreases geometrically

  • as you go down from level k, and also as you go up from level k.

  • So therefore we can just look at what the space is at this level

  • k here.

  • So the space is going to be p times the size of each one

  • of these nodes squared.

  • And the size of each one of these nodes

  • is going to be n over 2 to the log base 2 of p over 3.

  • And then we square that because we're using

  • n squared temporary space.

  • So if you solve that, that gives you p to the one-third times n

  • squared, which is better than the upper bound

  • we saw earlier of order p times n squared.

  • So you can work out the details for this example.

  • Not all the details are shown on this slide.

  • You need to show that the level k here actually dominates

  • all the other levels in the recursion tree.

  • But in general, if you know what the structure of the algorithm,

  • is you can potentially prove a stronger space bound than just

  • applying the general theorem we showed on the previous slide.

  • So any questions on this?

  • OK, so as I said before, the problem with heap-based linkage

  • is that parallel functions fail to interoperate

  • with legacy and third-party serial binaries.

  • Yes, was there a question?

  • STUDENT: I actually do have a question.

  • JULIAN SHUN: Yes.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yes.

  • STUDENT: How do we know that the workers don't split

  • along the path of the [INAUDIBLE] instead of across

  • or horizontal.

  • JULIAN SHUN: Yes.

  • So you don't actually know that.

  • But this turns out to be the worst case.

  • So if it branches any other way, the space

  • is just going to be lower.

  • So you have to argue that this is going to be the worst case,

  • and it's going to be--

  • intuitively it's the worst case, because you're

  • using most of the memory near the root of the recursion tree.

  • So if you can get all p nodes as close as possible to the root,

  • that's going to make your space as high as possible.

  • It's a good question.

  • So parallel functions fail to interoperate

  • with legacy and third-party serial binaries.

  • Even if you can recompile all of this code, which

  • isn't always necessarily the case,

  • you can still have issues if the legacy code

  • is taking advantage of the traditional linear stack

  • inside the source code.

  • So our implementation of cilk uses a less space

  • efficient strategy that is interoperable with legacy code.

  • And it uses a pool of linear stacks instead

  • of a heap-based strategy.

  • So we're going to maintain a pool of linear stacks lying

  • around.

  • There's going to be more than p stacks lying around.

  • And whenever a worker tries to steal something,

  • it's going to try to acquire one of these tasks

  • from this pool of linear tasks.

  • And when it's done, it will return it back.

  • But when it finds that there's no more

  • linear stacks in this pool, then it's

  • not going to steal anymore.

  • So this is still going to preserve the space bound,

  • as long as the number of stocks is a constant times the number

  • of processors.

  • But it will affect the time bounds

  • of the work-stealing algorithm.

  • Because now when a worker is idle,

  • it might not necessarily have the chance

  • to steal if there are no more stacks lying around.

  • This strategy doesn't require any changes

  • to the operating system.

  • There is a way where you can preserve the space

  • and the time bounds using thread local memory mapping.

  • But this does require changes to the operating system.

  • So our implementation of cilk uses a pool of linear stacks,

  • and it's based on the Intel implementation.

  • OK.

  • All right, so we talked about stacks,

  • and that we just reduce the problem to heap allocation.

  • So now we have to talk about heaps.

  • So let's review some basic properties

  • of heap-storage allocators.

  • So here's a definition.

  • The allocator speed is the number

  • of allocations and d allocations per second

  • that the allocator can sustain.

  • And here's a question.

  • Is it more important to maximize the allocator speed

  • for large blocks or small blocks?

  • Yeah?

  • STUDENT: Small blocks?

  • JULIAN SHUN: So small blocks.

  • Here's another question.

  • Why?

  • Yes?

  • STUDENT: So you're going to be doing a lot of [INAUDIBLE]..

  • JULIAN SHUN: Yes, so one answer is

  • that you're going to be doing a lot more allocations

  • and deallocations of small blocks than large blocks.

  • There's actually a more fundamental reason

  • why it's more important to optimize for small blocks.

  • So anybody?

  • Yeah?

  • STUDENT: [INAUDIBLE] basically not being

  • able to make use of pages.

  • JULIAN SHUN: Yeah, so that's another reason

  • for small blocks.

  • It's more likely that it will lead to fragmentation

  • if you don't optimize for small blocks.

  • What's another reason?

  • Yes.

  • STUDENT: Wouldn't it just take longer

  • to allocate larger blocks anyway?

  • So the overhead is going to be more noticeable if you have

  • a big overhead when you allocate small blocks

  • versus large blocks?

  • JULIAN SHUN: Yeah.

  • So the reason-- the main reason is that when you're allocating

  • a large--

  • when you're allocating a block, a user program

  • is typically going to write to all the bytes in the block.

  • And therefore, for a large block,

  • it takes so much time to write that the allocator

  • time has little effect on the overall running time.

  • Whereas if a program allocates many small blocks,

  • the amount of work-- useful work--

  • it's actually doing on the block is going to be--

  • it can be comparable to the overhead for the allocation.

  • And therefore, all of the allocation overhead

  • can add up to a significant amount for small blocks.

  • So essentially for large blocks, you

  • can amortize away the overheads for storage allocation,

  • whereas for small, small blocks, it's harder to do that.

  • Therefore, it's important to optimize for small blocks.

  • Here's another definition.

  • So the user footprint is the maximum

  • over time of the number u of bytes

  • in use by the user program.

  • And these are the bytes that are allocated and not freed.

  • And this is measuring the peak memory usage.

  • It's not necessarily equal to the sum of the sizes

  • that you have allocated so far, because you

  • might have reused some of that.

  • So the user footprint is the peak memory usage and number

  • of bytes.

  • And the allocator footprint is the maximum

  • over time of the number of a bytes

  • that the memory provided to the locator

  • by the operating system.

  • And the reason why the allocator footprint could be larger

  • than the user footprint, is that when

  • you ask the OS for some memory, it could give you

  • more than what you asked for.

  • And similarly, if you ask malloc for some amount of memory,

  • it can also give you more than what you asked for.

  • And the fragmentation is defined to be a divided by u.

  • And a program with low fragmentation

  • will keep this ratio as low as possible,

  • so keep the allocator footprint as close as

  • possible to the user footprint.

  • And in the best case, this ratio is going to be one.

  • So you're using all of the memory

  • that the operating system allocated.

  • One remark is that the allocator footprint

  • a usually gross monotonically for many allocators.

  • So it turns out that many allocators

  • do m maps to get more memory.

  • But they don't always free this memory back to the OS.

  • And you can actually free memory using something called

  • m unmap, which is the opposite of m map,

  • to give memory back to the OS.

  • But this turns out to be pretty expensive.

  • In modern operating systems, their implementation

  • is not very efficient.

  • So many allocators don't use m unmap.

  • You can also use something called m advise.

  • And what m advise does is it tells the operating system

  • that you're not going to be using this page anymore

  • but to keep it around in virtual memory.

  • So this has less overhead, because it

  • doesn't have to clear this entry from the page table.

  • It just has to mark that the program isn't

  • using this page anymore.

  • So some allocators use m advise with the option,

  • don't need, to free memory.

  • But a is usually still growing monotonically over time,

  • because allocators don't necessarily

  • free all of the things back to the OS that they allocated.

  • Here's a theorem that we proved in last week's lecture, which

  • says that the fragmentation for binned free list

  • is order log base 2 of u, or just order log u.

  • And the reason for this is that you're

  • can have log-based 2 of u bins.

  • And for each bin it can basically

  • contain u bytes of storage.

  • So overall you can use--

  • overall, you could have allocated

  • u times log u storage, and only be using u of those bytes.

  • So therefore the fragmentation is order log u.

  • Another thing to note is that modern 64-bit processors only

  • provide about 2 to 48 bytes of virtual address space.

  • So this is sort of news because you would probably

  • expect that, for a 64-bit processor,

  • you have to the 64 bytes of virtual address space.

  • But that turns out not to be the case.

  • So they only support to the 48 bytes.

  • And that turns out to be enough for all of the programs

  • that you would want to write.

  • And that's also going to be much more than the physical memory

  • you would have on a machine.

  • So nowadays, you can get a big server

  • with a terabyte of memory, or to the 40th bytes

  • of physical memory, which is still

  • much lower than the number of bytes in the virtual address

  • space.

  • Any questions?

  • OK, so here's some more definitions.

  • So the space overhead of an allocator

  • is a space used for bookkeeping.

  • So you could store--

  • perhaps you could store headers with the blocks

  • that you allocate to keep track of the size

  • and other information.

  • And that would contribute to the space overhead

  • Internal fragmentation is a waste

  • due to allocating larger blocks in the user request.

  • So you can get internal fragmentation

  • if, when you call malloc, you get back

  • a block that's actually larger than what the user requested.

  • We saw on the bin free list algorithm,

  • we're rounding up to the nearest power of 2's.

  • If you allocate nine bytes, you'll

  • actually get back 16 bytes in our binned-free list algorithm

  • from last lecture.

  • So that contributes to internal fragmentation.

  • It turns out that not all binned-free list

  • implementations use powers of 2.

  • So some of them use other powers that are smaller than 2

  • in order to reduce the internal fragmentation.

  • Then there's an external fragmentation,

  • which is the waste due to the inability to use storage

  • because it's not contiguous.

  • So for example, if I allocated a whole bunch of one byte things

  • consecutively in memory, then I freed every other byte.

  • And now I want to allocate a 2-byte thing,

  • I don't actually have contiguous mammary to satisfy that

  • request, because all of my free memory--

  • all of my free bytes are in one-bite chunks,

  • and they're not next to each other.

  • So this is one example of how external fragmentation can

  • happen after you allocate stuff and free stuff.

  • Then there's blow up.

  • And this is for a parallel locator.

  • The additional space beyond what a serial locator would require.

  • So if a serial locator requires s space,

  • and a parallel allocator requires t space,

  • then it's just going to be t over s.

  • That's the blow up.

  • OK, so now let's look at some parallel heap allocation

  • strategies.

  • So the first strategy is to use a global heap.

  • And this is how the default c allocator works.

  • So if you just use a default c allocator out of the box,

  • this is how it's implemented.

  • It uses a global heap where all the accesses

  • to this global heap are protected by mutex.

  • You can also use lock-free synchronization primitives

  • to implement this.

  • We'll actually talk about some of these synchronization

  • primitives later on in the semester.

  • And this is done to preserve atomicity

  • because you can have multiple threads trying

  • to access the global heap at the same time.

  • And you need to ensure that races are handled correctly.

  • So what's the blow up for this strategy?

  • How much more space am I using than just a serial allocator?

  • Yeah.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yeah, so the blow up is one.

  • Because I'm not actually using any more space

  • than the serial allocator.

  • Since I'm just maintaining one global heap, and everybody

  • is going to that heap to do allocations and deallocations.

  • But what's the potential issue with this approach?

  • Yeah?

  • STUDENT: Performance hit for that block coordination.

  • JULIAN SHUN: Yeah, so you're going

  • to take a performance hit for trying to acquire this lock.

  • So basically every time you do a allocation or deallocation,

  • you have to acquire this lock.

  • And this is pretty slow, and it gets

  • slower as you increase the number of processors.

  • Roughly speaking, acquiring a lock to perform

  • is similar to an L2 cache access.

  • And if you just run a serial allocator,

  • many of your requests are going to be satisfied just

  • by going into the L1 cache.

  • Because you're going to be allocating

  • things that you recently freed, and those things

  • are going to be residing in L1 cache.

  • But here, before you even get started,

  • you have to grab a lock.

  • And you have to pay a performance hit

  • similar to an L2 cache access.

  • So that's bad.

  • And it gets worse as you increase

  • the number of processors.

  • So the contention increases as you

  • increase the number of threads.

  • And then you can't--

  • you're not going to be able to get good scalability.

  • So ideally, as the number of threads or processors grows,

  • the time to perform an allocation or deallocation

  • shouldn't increase.

  • But in fact, it does.

  • And the most common reason for loss of scalability

  • is lock contention.

  • So here all of the processes are trying

  • to acquire the same lock, which is the same memory address.

  • And if you recall from the caching lecture,

  • or the multicore programming lecture,

  • every time you acquire a memory location,

  • you have to bring that cache line into your own cache,

  • and then invalidate the same cache line

  • in other processors' caches.

  • So if all the processors are doing this,

  • then this cache line is going to be bouncing around

  • among all of the processors' caches,

  • and this could lead to very bad performance.

  • So here's a question.

  • Is lock contention more of a problem for large blocks

  • or small blocks?

  • Yes.

  • STUDENT: So small blocks.

  • JULIAN SHUN: Here's another question.

  • Why?

  • Yes.

  • STUDENT: Because by the time it takes

  • to finish using the small block, then

  • the allocator is usually small.

  • So you do many allocations and deallocations,

  • which means you have to go through the lock multiple

  • times.

  • JULIAN SHUN: Yeah.

  • So one of the reasons is that when

  • you're doing small allocations, that

  • means that your request rate is going to be pretty high.

  • And your processors are going to be spending a lot of time

  • acquiring this lock.

  • And this can exacerbate the lock contention.

  • And another reason is that when you allocate a large block,

  • you're doing a lot of work, because you have to write--

  • most of the time you're going to write to all

  • the bytes in that large block.

  • And therefore you can amortize the overheads

  • of the storage allocator across all of the work

  • that you're doing.

  • Whereas for small blocks, in addition to

  • increasing this rate of memory requests, it's also--

  • there's much less work to amortized to overheads across.

  • So any questions?

  • OK, good.

  • All right.

  • So here's another strategy, which is to use local heaps.

  • So each thread is going to maintain its own heap.

  • And it's going to allocate out of its own heap.

  • And there's no locking that's necessary.

  • So when you allocate something, you get it from your own heap.

  • And when you free something, you put it back into your own heap.

  • So there's no synchronization required.

  • So that's a good thing.

  • It's very fast.

  • What's a potential issue with this approach?

  • Yes.

  • STUDENT: It's using a lot of extra space.

  • JULIAN SHUN: Yes, so this approach,

  • you're going to be using a lot of extra space.

  • So first of all, because you have

  • to maintain multiple heaps.

  • And what's one phenomenon that you

  • might see if you're executing a program

  • with this local-heap approach?

  • So it's a space--

  • could the space potentially keep growing over time?

  • Yes.

  • STUDENT: You could maybe like allocate

  • every one process [INAUDIBLE].

  • JULIAN SHUN: Yeah.

  • Yeah, so you could actually have an unbounded blow up.

  • Because if you do all of the allocations in one heap,

  • and you free everything in another heap,

  • then whenever the first heap does an allocation,

  • there's actually free space sitting around in another heap.

  • But it's just going to grab more memory from the operating

  • system.

  • So you're blow up can be unbounded.

  • And this phenomenon, it's what's called memory drift.

  • So blocks allocated by one thread

  • are freed by another thread.

  • And if you run your program for long enough,

  • your memory consumption can keep increasing.

  • And this is sort of like a memory leak.

  • So you might see that if you have a memory drift problem,

  • your program running on multiple processors

  • could run out of memory eventually.

  • Whereas if you just run it on a single core,

  • it won't run out of memory.

  • And here it's because the allocator isn't smart enough

  • to reuse things in other heaps.

  • So what's another strategy you can use to try to fix this?

  • Yes?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Sorry, can you repeat your question?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Because if you keep allocating

  • from one thread, if you do all of your allocations

  • in one thread, and do all of your deallocations

  • on another thread, every time you

  • allocate from the first thread, there's

  • actually memory sitting around in the system.

  • But the first thread isn't going to see it, because it only

  • sees its own heap.

  • And it's just going to keep grabbing

  • more memory from the OS.

  • And then the second thread actually

  • has this extra memory sitting around.

  • But it's not using it.

  • Because it's only doing the freeze.

  • It's not doing allocate.

  • And if we recall the definition of blow up

  • is, how much more space you're using

  • compared to a serial execution of a program.

  • If you executed this program on a single core,

  • you would only have a single heap that does the allocations

  • and frees.

  • So you're not going to--

  • your memory isn't going to blow up.

  • It's just going to be constant over time.

  • Whereas if you use two threads to execute this,

  • the memory could just keep growing over time.

  • Yes?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: So, it just--

  • so if you remember the binned-free list approach,

  • let's say we're using that.

  • Then all you have to do is set some pointers

  • in your binned-free lists data structure,

  • as well as the block that you're freeing,

  • so that it appears in one of the linked lists.

  • So you can do that even if some other processor allocated

  • that block.

  • OK, so what what's another strategy that can avoid

  • this issue of memory drift?

  • Yes?

  • STUDENT: Periodically shuffle the free memory that's

  • being used on different heaps.

  • JULIAN SHUN: Yeah.

  • So that's a good idea.

  • You could periodically rebalance the memory.

  • What's a simpler approach to solve this problem?

  • Yes?

  • STUDENT: Make it all know all of the free memory?

  • JULIAN SHUN: Sorry, could you repeat that?

  • STUDENT: Make them all know all of the free memory?

  • JULIAN SHUN: Yes.

  • So you could have all of the processors

  • know all the free memory.

  • And then every time it grabs something,

  • it looks in all the other heaps.

  • That does require a lot of synchronization overhead.

  • Might not perform that well.

  • What's an easier way to solve this problem?

  • Yes.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: So you could restructure your program

  • so that the same thread does the allocation

  • and frees for the same memory block.

  • But what if you didn't want to restructure your program?

  • How can you change the allocator?

  • So we want the behavior that you said,

  • but we don't want to change our program.

  • Yes.

  • STUDENT: You could have a single free list that's

  • protected by synchronization.

  • JULIAN SHUN: Yeah, so you could have a single free list.

  • But that gets back to the first strategy

  • of having a global heap.

  • And then you have high synchronization overheads.

  • Yes.

  • STUDENT: You could have the free map to the thread

  • that it came from or for the pointer that corresponds to--

  • that allocated it.

  • JULIAN SHUN: So you're saying free back

  • to the thread that allocated it?

  • Yes, so that that's exactly right.

  • So here each object, when you allocate it,

  • it's labeled with an owner.

  • And then whenever you free it, you

  • return it back to the owner.

  • So the objects that are allocated

  • will eventually go back to the owner's heap

  • if they're not in use.

  • And they're not going to be free lying around

  • in somebody else's heap.

  • The advantage of this approach is

  • that you get fast allocation and freeing of local objects.

  • Local objects are objects that you allocated.

  • However, free remote objects require some synchronization.

  • Because you have to coordinate with the other threads' heap

  • that you're sending the memory object back to.

  • But this synchronization isn't as bad as having a global heap,

  • since you only have to talk to one other thread in this case.

  • You can also bound the blow up by p.

  • So the reason why the blow up is upper bounded by p

  • is that, let's say the serial allocator uses

  • at most x memory.

  • In this case, each of the heaps can use at most x memory,

  • because that's how much the serial program would have used.

  • And you have p of these heaps, so overall you're

  • using p times x memory.

  • And therefore the ratio is upper bounded by p.

  • Yes?

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: So when you free an object, it goes--

  • if you allocated that object, it goes back to your own heap.

  • If your heap is empty, it's actually

  • going to get more memory from the operating system.

  • It's not going to take something from another thread's heap.

  • But the maximum amount of memory that you're going to allocate

  • is going to be upper bounded by x.

  • Because the sequential serial program took that much.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Yeah.

  • So the upper bound for the blow up is p.

  • Another advantage of this approach

  • is that it's resilience--

  • it has resilience to false sharing.

  • So let me just talk a little bit about false sharing.

  • So true sharing is when two processors

  • are trying to access the same memory location.

  • And false sharing is when multiple processors are

  • accessing different memory locations,

  • but those locations happen to be on the same cache line.

  • So here's an example.

  • Let's say we have two variables, x and y.

  • And the compiler happens to place x and y on the same cache

  • line.

  • Now, when the first processor writes to x,

  • it's going to bring this cache line into its cache.

  • When the other processor writes to y,

  • since it's on the same cache line,

  • it's going to bring this cache line to y's cache.

  • And then now, the first processor writes x,

  • it's going to bring this cache line back

  • to the first processor's cache.

  • And then you can keep--

  • you can see this phenomenon keep happening.

  • So here, even though the processors

  • are writing to different memory locations,

  • because they happen to be on the same cache line,

  • the cache line is going to be bouncing back and forth

  • on the machine between the different processors' caches.

  • And this problem gets worse if more processors

  • are accessing this cache line.

  • So in this-- this can be quite hard to debug.

  • Because if you're using just variables on the stack,

  • you don't actually know necessarily

  • where the compiler is going to place these memory locations.

  • So the compiler could just happen

  • to place x and y in the same cache block.

  • And then you'll get this performance hit,

  • even though it seems like you're accessing different memory

  • locations.

  • If you're using the heap for memory allocation,

  • you have more knowledge.

  • Because if you allocate a huge block,

  • you know that all of the memory locations

  • are contiguous in physical memory.

  • So you can just space your--

  • you can space the accesses far enough apart so

  • that different processes aren't going

  • to touch the same cache line.

  • A more general approach is that you can actually

  • pad the object.

  • So first, you can align the object

  • on a cache line boundary.

  • And then you pad out the remaining memory locations

  • of the objects so that it fills up the entire cache line.

  • And now there's only one thing on that cache line.

  • But this does lead to a waste of space

  • because you have this wasted padding here.

  • So program can induce false sharing

  • by having different threads process

  • nearby objects, both on the stack and on the heap.

  • And then an allocator can also induce false sharing

  • in two ways.

  • So it can actively induce false sharing.

  • And this is when the allocator satisfies memory requests

  • from different threads using the same cache block.

  • And it can also do this passively.

  • And this is when the program passes objects lying around

  • in the same cache line.

  • So different threads, and then the allocator

  • reuses the object storage after the objects

  • are free to satisfy requests from those different threads.

  • And the local ownership approach tends

  • to reduce false sharing because the thread that

  • allocates an object is eventually

  • going to get it back.

  • You're not going to have it so that an object is permanently

  • split among multiple processors' heaps.

  • So even if you see false sharing in local ownership,

  • it's usually temporary.

  • Eventually it's going-- the object is

  • going to go back to the heap that it was allocated from,

  • and the false sharing is going to go away.

  • Yes?

  • STUDENT: Are the local heaps just three to five regions in

  • [INAUDIBLE]?

  • JULIAN SHUN: I mean, you can implement it in various ways.

  • I mean can have each one of them have a binned-free list

  • allocator, so there's no restriction

  • on where they have to appear in physical memory.

  • There are many different ways where you can--

  • you can basically plug-in any serial locator

  • for the local heap.

  • So let's go back to parallel heap allocation.

  • So I talked about three approaches already.

  • Here's a fourth approach.

  • This is called the hoard allocator.

  • And this was actually a pretty good allocator

  • when it was introduced almost two decades ago.

  • And it's inspired a lot of further research

  • on how to improve parallel-memory allocation.

  • So let me talk about how this works.

  • So in the hoard allocator, we're going to have p local heaps.

  • But we're also going to have a global heap.

  • The memory is going to be organized

  • into large super blocks of size s.

  • And s is usually a multiple of the page size.

  • So this is the granularity at which

  • objects are going to be moved around in the allocator.

  • And then you can move super blocks between the local heaps

  • and the global heaps.

  • So when a local heap becomes--

  • has a lot of super blocks that are not being fully used

  • and you can move it to the global heap,

  • and then when a local heap doesn't have enough memory,

  • it can go to the global heap to get more memory.

  • And then when the global heap doesn't have any more memory,

  • then it gets more memory from the operating system.

  • So this is sort of a combination of the approaches

  • that we saw before.

  • The advantages are that this is a pretty fast allocator.

  • It's also scalable.

  • As you add more processors, the performance improves.

  • You can also bound the blow up.

  • And it also has resilience to false sharing,

  • because it's using local heaps.

  • So let's look at how an allocation using the hoard

  • allocator works.

  • So let's just assume without loss of generality

  • that all the blocks are the same size.

  • So we have fixed-size allocation.

  • So let's say we call malloc in our program.

  • And let's say thread i calls the malloc.

  • So what we're going to do is we're

  • going to check if there is a free object in heap i

  • that can satisfy this request.

  • And if so, we're going to get an object

  • from the fullest non-full super block in i's heap.

  • Does anyone know why we want to get the object from the fullest

  • non-full super block?

  • Yes.

  • STUDENT: [INAUDIBLE]

  • JULIAN SHUN: Right.

  • So when a super block needs to be moved,

  • it's as dense as possible.

  • And more importantly, this is to reduce external fragmentation.

  • Because as we saw in the last lecture,

  • if you skew the distribution of allocated memory objects

  • to as few pages, or in this case,

  • as few super blocks as possible, that

  • reduces your external fragmentation.

  • OK, so if it finds it in its own heap,

  • then it's going to allocate an object from there.

  • Otherwise, it's going to check the global heap.

  • And if there's something in the global heap--

  • so here it says, if the global heap is empty,

  • then it's going to get a new super block from the OS.

  • Otherwise, we can get a super block from the global heap,

  • and then use that one.

  • And then finally we set the owner

  • of the block we got either from the OS or from the global heap

  • to i, and then we return that free object to the program.

  • So this is how a malloc works using the hoard allocator.

  • And now let's look at hoard deallocation.

  • Let use of i be the in use storage in heap i.

  • This is the heap for thread i.

  • And let a sub i be the storage owned by heap i.

  • The hoard allocator maintains the following invariant

  • for all heaps i.

  • And the invariant is as follows.

  • So u sub i is always going to be greater

  • than or equal to the min of a sub i minus 2 times s.

  • Recall s is the super block size.

  • And a sub i over 2.

  • So how it implements this is as follows.

  • When we call free of x, let's say x is owned by thread i,

  • then we're going to put x back into heap i,

  • and then we're going to check if the n u storage in heap i,

  • u sub i is less than the min of a sub i minus 2 s

  • and a sub i over 2.

  • And what this condition says, if it's true,

  • it means that your heap is, at most, half utilized.

  • Because if it's smaller than this,

  • it has to be smaller than a sub i over 2.

  • That means there's twice as much allocated

  • than used in the local heap i.

  • And therefore there must be some super block

  • that's at least half empty.

  • And you move that super block, or one of those super blocks,

  • to the global heap.

  • So any questions on how the allocation and deallocation

  • works?

  • So since we're maintaining this invariant,

  • it's going to allow us to approve a bound on the blow up.

  • And I'll show you that on the next slide.

  • But before I go on, are there any questions?

  • OK, so let's look at how we can bound the blow up

  • of the hoard allocator.

  • So there is actually a lemma that we're

  • going to use and not prove.

  • The lemma is that the maximum storage allocated

  • in the global heap is at most a maximum storage allocated

  • in the local heaps.

  • So we just need to analyze how much storage is

  • allocated in the local heaps.

  • Because the total amount of storage

  • is going to be, at most, twice as much,

  • since the global heap storage is dominated by the local heap

  • storage.

  • So you can prove this lemma by case analysis.

  • And there's the hoard paper that's

  • available on learning modules.

  • And you're free to look at that if you want

  • to look at how this is proved.

  • But here I'm just going to use this lemma

  • to prove this theorem, which says that, let u be the user

  • footprint for a program.

  • And let a be the hoard's allocator footprint.

  • We have that a as upper bounded by order u plus s p.

  • And therefore, a divided by u, which is a blowup,

  • is going to be 1 plus order s p divided by u.

  • OK, so let's see how this proof works.

  • So we're just going to analyze the storage in the local heaps.

  • Now recall that we're always satisfying this invariant here,

  • where u sub i is greater than the min of a sub i minus 2 s

  • and a sub i over 2.

  • So the first term says that we can

  • have 2 s on utilized storage per heap.

  • So it's basically giving two super blocks for free

  • to each heap.

  • And they don't have to use it.

  • They can basically use it as much as they want.

  • And therefore, the total amount of storage contributed

  • by the first term is going to be order

  • s p, because each processor has up to 2 s unutilized storage.

  • So that's where the second term comes from here.

  • And the second term, a sub i over 2--

  • this will give us the first-term order u.

  • So this says that the allocated storage

  • is at most twice the use storage for--

  • and then if you sum up across all the processors,

  • then there's a total of order use storage that's allocated.

  • Because the allocated storage can be at most

  • twice the used storage.

  • OK, so that's the proof of the blow up for hoard.

  • And this is pretty good.

  • It's 1 plus some lower order term.

  • OK, so-- now these are some other allocators

  • that people use.

  • So jemalloc is a pretty popular one.

  • Has a few differences with hoard.

  • It has a separate global lock for each different allocation

  • size.

  • It allocates the object with the smallest address

  • among all the objects of the requested size.

  • And it releases empty pages using m advise,

  • which we talked about--

  • I talked about earlier.

  • And it's pretty popular because it has good performance,

  • and it's pretty robust to different allocation traces.

  • There's also another one called SuperMalloc, which

  • is an up and coming contender.

  • And it was developed by Bradley Kuszmaul.

  • Here are some allocator speeds for the allocators

  • that we looked at for our particular benchmark.

  • And for this particular benchmark,

  • we can see that SuperMalloc actually does really well.

  • It's more than three times faster than jemalloc,

  • and jemalloc is more than twice as fast as hoard.

  • And then the default allocator, which

  • uses a global heap is pretty slow, because it

  • can't get good speed up.

  • And all these experiments are in 32 threads.

  • I also have the lines of code.

  • So we see that SuperMalloc actually

  • has very few lines of code compared

  • to the other allocators.

  • So it's relatively simple.

  • OK so, I also have some slides in Garbage Collection.

  • But since we're out of time, I'll just

  • put these slides online and you can read them.

VOICEOVER: The following content is provided under a Creative

字幕與單字

單字即點即查 點擊單字可以查詢單字解釋

B1 中級

12.並行存儲分配 (12. Parallel Storage Allocation)

  • 4 0
    林宜悉 發佈於 2021 年 01 月 14 日
影片單字