In CUDA SDK function cudaHostAlloc() for allocation new memory region can use flags:
cudaHostAllocDefault (default - 0 and causes cudaHostAlloc() to emulate cudaMallocHost())
cudaHostAllocPortable
cudaHostAllocMapped
cudaHostAllocWriteCombined
To mark memory region that already allocated we can use cudaHostRegister() with flags:
0 (default)
cudaHostRegisterPortable
cudaHostRegisterMapped
Why we can mark memory WriteCombined when allocating it by flag cudaHostAllocWriteCombined by using cudaHostAlloc(), but can't mark as WriteCombined already existing memory region by using cudaHostRegister()?
Already allocated memory we must will mark only through the POSIX function set_memory_wc()?
I did not know of any APIs that could change the cacheability of an existing VA range until you referenced set_memory_wc(). Such an operation would be extremely expensive due to all the cache flushes and TLB shootdowns that would be required; and the memory would basically be unreadable until you found some way to unmark it as WC.
Why are you trying to use WC memory? On pre-i7 (Nehalem) CPUs, WC had slightly higher transfer performance (IIRC) because it inhibited snooping of PCI Express traffic to and from the memory. But on Nehalem and later CPUs, I don't know of any application that has concretely demonstrated a benefit from WC memory.
Related
Supose linux-32: the aligment rules say, for example, that doubles (8 Bytes) must be aligned to 4 Bytes. This means that, if we assume 64 Bytes cache blocks (a typical value for modern processors) we can have a double aligned in the 60th position, which mean that this double will be in 2 different cache blocks.
It could even happen that both parts of the double were in 2 different cache blocks located in 2 different 4KB pages.
After this brief introduction to put the question in context, I have a couple of doubts:
1- For an assembler programming where we seek maximum performance, it is recommended to prevent these things from happenning by putting alignment directives, right? Or, for any reason that I unknow, making the alignment to make the double in only 1 block doesn't imply any performance change?
2- How will be the store instruction decoded in the in the mentioned case? (supose modern intel microarchitecture). I mean, I know that a normal store x86 instruction is decoded in a micro-fused pair of str-addr and str-data, but in this case where 2 different cache blocks (and maybe even 2 different 4KB pages) are involved, this will be decoded in 2 micro-fused pair of str-addr and str-data (one for the first 4 bytes of the double and another for the last 4 bytes)? Or it will be decoded to a single micro-fused pair but having to do both the str-addr and the str-data twice the work until finally being able to exit the execution port?
Yes, of course you should align a double whenever possible, like compilers do except when forced by ABI struct-layout rules to misalign them. (The ABI was designed when i386 was current so a double always required 2 loads anyway.)
The current version of the i386 System V ABI requires 16-byte stack alignment, so local doubles (that have to get spilled at all instead of kept in regs) can be aligned, and malloc has to return memory suitable for any type, and alignof(max_align_t) = 16 on 32-bit Linux (8 on 32-bit Windows) so 32-bit malloc will always give you at least 16 (or 8)-byte aligned memory. And of course in static storage you control the alignment with align (NASM) or .p2align (GAS) directives.
For the perf downsides of cacheline splits and page splits, see How can I accurately benchmark unaligned access speed on x86_64
re: decoding: The address isn't know at decode time so obviously any effects of a line-split page-split are resolved later. For stores, probably no effect until the store-buffer entry has to commit to L1d cache. Are two store buffer entries needed for split line/page stores on recent Intel? - probably no, allocating a 2nd entry after executing the store-address uop is implausible.
For loads, re-running the load through the execution unit to get the other half (or whatever uneven split), using internal line-split buffers to combine data. (Not re-dispatching from the RS, just internally handled in the load port. But the RS does aggressively replay uops waiting for the result of a load.)
Re-running the store-data uop for a misaligned store seems unlikely, too. I don't think we see extra counts for uops_dispatched_port.port_4 perf events.
I understand that in Linux the mm_struct describes the memory layout of a process. I also understand that the start_brk and brk mark the start and end of the heap section of a process respectively.
Now, this is my problem: I have a process, for which I wrote the source code, that allocates 5.25 GB of heap memory using malloc. However, when I examine the process's mm_sruct using a kernel module I find the value of is equal to 135168. And this is different from what I expected: I expected brk - start_brk to be equal slight above 5.25 GB.
So, what is going on here?
Thanks.
I notice the following in the manpage for malloc(3):
Normally, malloc() allocates memory from the heap, and adjusts the size of the heap as required, using sbrk(2). When allocating blocks of memory larger than MMAP_THRESHOLD bytes, the glibc malloc() implementation allocates the memory as a private anonymous mapping using mmap(2). MMAP_THRESHOLD is 128 kB by default, but is adjustable using mallopt(3). Allocations performed using mmap(2) are unaffected by the RLIMIT_DATA resource limit (see getrlimit(2)).
So it sounds like mmap is used instead of the heap.
I have a process will do much lithography calculation, so I used mmap to alloc some memory for memory pool. When process need a large chunk of memory, I used mmap to alloc a chunk, after use it then put it in the memory pool, if the same chunk memory is needed again in the process, get it from the pool directly, not used memory map again.(not alloc all the need memory and put it in the pool at the beginning of the process). Between mmaps function, there are some memory malloc not used mmap, such as malloc() or new().
Now the question is:
If I used memset() to set all the chunk data to ZERO before putting it in the memory pool, the process will use too much virtual memory as following, format is "mmap(size)=virtual address":
mmap(4198400)=0x2aaab4007000
mmap(4198400)=0x2aaab940c000
mmap(8392704)=0x2aaabd80f000
mmap(8392704)=0x2aaad6883000
mmap(67112960)=0x2aaad7084000
mmap(8392704)=0x2aaadb085000
mmap(2101248)=0x2aaadb886000
mmap(8392704)=0x2aaadba89000
mmap(67112960)=0x2aaadc28a000
mmap(2101248)=0x2aaae028b000
mmap(2101248)=0x2aaae0c8d000
mmap(2101248)=0x2aaae0e8e000
mmap(8392704)=0x2aaae108f000
mmap(8392704)=0x2aaae1890000
mmap(4198400)=0x2aaae2091000
mmap(4198400)=0x2aaae6494000
mmap(8392704)=0x2aaaea897000
mmap(8392704)=0x2aaaeb098000
mmap(2101248)=0x2aaaeb899000
mmap(8392704)=0x2aaaeba9a000
mmap(2101248)=0x2aaaeca9c000
mmap(8392704)=0x2aaaec29b000
mmap(8392704)=0x2aaaecc9d000
mmap(2101248)=0x2aaaed49e000
mmap(8392704)=0x2aaafd6a7000
mmap(2101248)=0x2aacc5f8c000
The mmap last - first = 0x2aacc5f8c000 - 0x2aaab4007000 = 8.28G
But if I don't call memset before put in the memory pool:
mmap(4198400)=0x2aaab4007000
mmap(8392704)=0x2aaab940c000
mmap(8392704)=0x2aaad2480000
mmap(67112960)=0x2aaad2c81000
mmap(2101248)=0x2aaad6c82000
mmap(4198400)=0x2aaad6e83000
mmap(8392704)=0x2aaadb288000
mmap(8392704)=0x2aaadba89000
mmap(67112960)=0x2aaadc28a000
mmap(2101248)=0x2aaae0a8c000
mmap(2101248)=0x2aaae0c8d000
mmap(2101248)=0x2aaae0e8e000
mmap(8392704)=0x2aaae1890000
mmap(8392704)=0x2aaae108f000
mmap(4198400)=0x2aaae2091000
mmap(4198400)=0x2aaae6494000
mmap(8392704)=0x2aaaea897000
mmap(8392704)=0x2aaaeb098000
mmap(2101248)=0x2aaaeb899000
mmap(8392704)=0x2aaaeba9a000
mmap(2101248)=0x2aaaec29b000
mmap(8392704)=0x2aaaec49c000
mmap(8392704)=0x2aaaecc9d000
mmap(2101248)=0x2aaaed49e000
The mmap last - first = 0x2aaaed49e000 - 0x2aaab4007000= 916M
So the first process will "out of memory" and killed.
In the process, the mmap memory chunk will not be fully used or not even used although it is alloced, I mean, for example, before calibration, the process mmap 67112960(64M), it will not used(write or read data in this memory region) or just used the first 2M bytes, then put in the memory pool.
I know the mmap just return virtual address, the physical memory used delay alloc, it will be alloced when read or write on these address.
But what made me confused is that, why the virtual address increase so much? I used the centos 5.3, kernel version is 2.6.18, I tried this process both on libhoard and the GLIBC(ptmalloc), both with the same behavior.
Do anyone meet the same issue before, what is the possible root cause?
Thanks.
VMAs (virtual memory areas, AKA memory mappings) do not need to be contiguous. Your first example uses ~256 Mb, the second ~246 Mb.
Common malloc() implementations use mmap() automatically for large allocations (usually larger than 64Kb), freeing the corresponding chunks with munmap(). So you do not need to mmap() manually for large allocations, your malloc() library will take care of that.
When mmap()ing, the kernel returns a COW copy of a special zero page, so it doesn't allocate memory until it's written to. Your zeroing is causing memory to be really allocated, better just return it to the allocator, and request a new memory chunk when you need it.
Conclusion: don't write your own memory management unless the system one has proven inadecuate for your needs, and then use your own memory management only when you have proved it noticeably better for your needs with real life load.
I want to allocate a large DMA buffer, about 40 MB in size. When I use dma_alloc_coherent(), it fails and what I see is:
------------[ cut here ]------------
WARNING: at mm/page_alloc.c:2106 __alloc_pages_nodemask+0x1dc/0x788()
Modules linked in:
[<8004799c>] (unwind_backtrace+0x0/0xf8) from [<80078ae4>] (warn_slowpath_common+0x4c/0x64)
[<80078ae4>] (warn_slowpath_common+0x4c/0x64) from [<80078b18>] (warn_slowpath_null+0x1c/0x24)
[<80078b18>] (warn_slowpath_null+0x1c/0x24) from [<800dfbd0>] (__alloc_pages_nodemask+0x1dc/0x788)
[<800dfbd0>] (__alloc_pages_nodemask+0x1dc/0x788) from [<8004a880>] (__dma_alloc+0xa4/0x2fc)
[<8004a880>] (__dma_alloc+0xa4/0x2fc) from [<8004b0b4>] (dma_alloc_coherent+0x54/0x60)
[<8004b0b4>] (dma_alloc_coherent+0x54/0x60) from [<803ced70>] (mxc_ipu_ioctl+0x270/0x3ec)
[<803ced70>] (mxc_ipu_ioctl+0x270/0x3ec) from [<80123b78>] (do_vfs_ioctl+0x80/0x54c)
[<80123b78>] (do_vfs_ioctl+0x80/0x54c) from [<8012407c>] (sys_ioctl+0x38/0x5c)
[<8012407c>] (sys_ioctl+0x38/0x5c) from [<80041f80>] (ret_fast_syscall+0x0/0x30)
---[ end trace 4e0c10ffc7ffc0d8 ]---
I've tried different values and it looks like dma_alloc_coherent() can't allocate more than 2^25 bytes (32 MB).
How can such a large DMA buffer can be allocated?
After the system has booted up dma_alloc_coherent() is not necessarily reliable for large allocations. This is simply because non-moveable pages quickly fill up your physical memory making large contiguous ranges rare. This has been a problem for a long time.
Conveniently a recent patch-set may help you out, this is the contiguous memory allocator which appeared in kernel 3.5. If you're using a kernel with this then you should be able to pass cma=64M on your kernel command line and that much memory will be reserved (only moveable pages will be placed there). When you subsequently ask for your 40M allocation it should reliably succeed. Simples!
For more information check out this LWN article:
https://lwn.net/Articles/486301/
I am trying to understand how heap fragmenation works. What does the following output tell me?
Is this heap overly fragmented?
I have 243010 "free objects" with a total of 53304764 bytes. Are those "free object" spaces in the heap that once contained object but that are now garabage collected?
How can I force a fragmented heap to clean up?
!dumpheap -type Free -stat
total 243233 objects
Statistics:
MT Count TotalSize Class Name
0017d8b0 243010 53304764 Free
It depends on how your heap is organized. You should have a look at how much memory in Gen 0,1,2 is allocated and how much free memory you have there compared to the total used memory.
If you have 500 MB managed heap used but and 50 MB is free then you are doing pretty well. If you do memory intensive operations like creating many WPF controls and releasing them you need a lot more memory for a short time but .NET does not give the memory back to the OS once you allocated it. The GC tries to recognize allocation patterns and tends to keep your memory footprint high although your current heap size is way too big until your machine is running low on physical memory.
I found it much easier to use psscor2 for .NET 3.5 which has some cool commands like ListNearObj where you can find out which objects are around your memory holes (pinned objects?). With the commands from psscor2 you have much better chances to find out what is really going on in your heaps. Most commands are also available in SOS.dll in .NET 4 as well.
To answer your original question: Yes free objects are gaps on the managed heap which can simply be the free memory block after your last allocated object on a GC segement. Or if you do !DumpHeap with the start address of a GC segment you see the objects allocated in that managed heap segment along with your free objects which are GC collected objects.
This memory holes do normally happen in Gen2. The object addresses before and after the free object do tell you what potentially pinned objects are around your hole. From this you should be able to determine your allocation history and optimize it if you need to.
You can find the addresses of the GC Heaps with
0:021> !EEHeap -gc
Number of GC Heaps: 1
generation 0 starts at 0x101da9cc
generation 1 starts at 0x10061000
generation 2 starts at 0x02aa1000
ephemeral segment allocation context: none
segment begin allocated size
02aa0000 02aa1000** 03836a30 0xd95a30(14244400)
10060000 10061000** 103b8ff4 0x357ff4(3506164)
Large object heap starts at 0x03aa1000
segment begin allocated size
03aa0000 03aa1000 03b096f8 0x686f8(427768)
Total Size: Size: 0x115611c (18178332) bytes.
------------------------------
GC Heap Size: Size: 0x115611c (18178332) bytes.
There you see that you have heaps at 02aa1000 and 10061000.
With !DumpHeap 02aa1000 03836a30 you can dump the GC Heap segment.
!DumpHeap 02aa1000 03836a30
Address MT Size
...
037b7b88 5b408350 56
037b7bc0 60876d60 32
037b7be0 5b40838c 20
037b7bf4 5b408350 56
037b7c2c 5b408728 20
037b7c40 5fe4506c 16
037b7c50 60876d60 32
037b7c70 5b408728 20
037b7c84 5fe4506c 16
037b7c94 00135de8 519112 Free
0383685c 5b408728 20
03836870 5fe4506c 16
03836880 608c55b4 96
....
There you find your free memory blocks which was an object which was already GCed. You can dump the surrounding objects (the output is sorted address wise) to find out if they are pinned or have other unusual properties.
You have 50MB of RAM as Free space. This is not good.
Having .NET allocating blocks of 16MB from process, we have a fragmentation issue indeed.
There are plenty of reasons to fragmentation to occure in .NET.
Have a look here and here.
In your case it is possibly a pinning. As 53304764 / 243010 makes 219.35 bytes per object - much lower then LOH objects.