为什么memset会变慢？

Question

我的CPU的规格说它应该为内存带来5.336GB/s的带宽.为了测试这个,我写了一个简单的程序,在一个大数组上运行memset(或memcpy)并报告时间.我在memset上显示3.8GB/s,在memcpy上显示1.9GB/s. http://en.wikipedia.org/wiki/Intel_Core_(microarchitecture)说我的Q9400应该达到5.336MB/s.怎么了？

我试过用赋值循环替换memset或memcpy.我已经google了一下,试图了解内存对齐情况.我尝试过不同的编译器标志.我花了很多时间在这上面尴尬.感谢您的任何帮助,您可以提供!

我正在使用Ubuntu 12.04和libc-dev版本2.15-0ubuntu10.5和内核3.8.0-37-generic

代码:

#include <stdio.h>
#include <time.h>
#include <string.h>
#include <stdlib.h>

#define numBytes ((long)(1024*1024*1024))
#define numTransfers ((long)(8))

int main(int argc,char**argv){
    if(argc!=3){
        printf("Usage: %s BLOCK_SIZE_IN_BYTES NUMBER_OF_BLOCKS_TO_TRANSFER\n",argv[0]);
        return -1;
    }
    char*__restrict__ source=(char*)malloc(numBytes);
    char*__restrict__ dest=(char*)malloc(numBytes);
    struct timespec start,end;
    long totalTimeMs;
    int i;

    clock_gettime(CLOCK_MONOTONIC_RAW,&start);
    for(i=0;i<numTransfers;++i)
        memset(source,0,numBytes);
    clock_gettime(CLOCK_MONOTONIC_RAW,&end);
    totalTimeMs=(end.tv_nsec-start.tv_nsec)*.000001+1000*(end.tv_sec-start.tv_sec);
    printf("memset %ld bytes %ld times (%.2fGB total) in %ldms (%.3fGB/s). ",numBytes,numTransfers,numBytes/1024.0/1024/1024*numTransfers,totalTimeMs,numBytes/1024.0/1024/1024*1000*numTransfers/totalTimeMs);

    clock_gettime(CLOCK_MONOTONIC_RAW,&start);
    for(i=0;i<numTransfers;++i)
        memcpy( dest, source, numBytes);
    clock_gettime(CLOCK_MONOTONIC_RAW,&end);
    totalTimeMs=(end.tv_nsec-start.tv_nsec)*.000001+1000*(end.tv_sec-start.tv_sec);
    printf("memcpy %ld bytes %ld times (%.2fGB total) in %ldms (%.3fGB/s).\n",numBytes,numTransfers,numBytes/1024.0/1024/1024*numTransfers,totalTimeMs,numBytes/1024.0/1024/1024*1000*numTransfers/totalTimeMs);

    free(source);
    free(dest);

    return EXIT_SUCCESS;
}

编译命令:

gcc -O3 -DNDEBUG -o memcpyStackOverflowNoParameters.c.o -c memcpyStackOverflowNoParameters.c
gcc -O3 -DNDEBUG memcpyStackOverflowNoParameters.c.o -o memcpy -rdynamic -lrt 

样本输出:

memset 1073741824 bytes 8 times (8.00GB total) in 2214ms (3.880GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4466ms (1.923GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2218ms (3.873GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4557ms (1.885GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2222ms (3.866GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4433ms (1.938GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2216ms (3.876GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4521ms (1.900GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2217ms (3.875GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4520ms (1.900GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2218ms (3.873GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4430ms (1.939GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2226ms (3.859GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4444ms (1.933GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2225ms (3.861GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4485ms (1.915GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2620ms (3.279GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4855ms (1.769GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2535ms (3.389GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4870ms (1.764GB/s).
memset 1073741824 bytes 8 times (8.00GB total) in 2423ms (3.545GB/s). memcpy 1073741824 bytes 8 times (8.00GB total) in 4905ms (1.751GB/s).

我的硬件根据lshw:

  product: OptiPlex 960 ()
  vendor: Winbond Electronics
  width: 64 bits
*-core
     description: Motherboard
     product: 0Y958C
     vendor: Winbond Electronics
   *-firmware
        description: BIOS
        capabilities: pci pnp apm upgrade shadowing escd cdboot bootselect edd int13floppytoshiba int13floppy720 int5printscreen int9keyboard int14serial int17printer acpi usb biosbootspecification netboot
   *-cpu
        product: Intel(R) Core(TM)2 Quad CPU    Q9400  @ 2.66GHz
        physical id: 400
        size: 2666MHz
        width: 64 bits
        clock: 1333MHz
        capabilities: x86-64 fpu fpu_exception wp vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe syscall nx constant_tsc arch_perfmon pebs bts rep_good nopl aperfmperf pni dtes64 monitor ds_cpl vmx smx est tm2 ssse3 cx16 xtpr pdcm sse4_1 xsave lahf_lm dtherm tpr_shadow vnmi flexpriority
        configuration: cores=4 enabledcores=4 threads=4
      *-cache:0
           description: L1 cache
           physical id: 700
           size: 256KiB
           capacity: 256KiB
           capabilities: internal write-back unified
      *-cache:1
           description: L2 cache
           physical id: 701
           size: 6MiB
           capacity: 6MiB
           capabilities: internal varies unified
   *-memory
        description: System Memory
        physical id: 1000
        slot: System board or motherboard
        size: 4GiB
      *-bank:0
           description: DIMM DDR2 Synchronous 667 MHz (1.5 ns)
           product: CT51264AA667.M16FC
           vendor: 7F7F7F7F7F9B0000
           slot: DIMM_1
           size: 4GiB
           width: 64 bits
           clock: 667MHz (1.5ns)
      *-bank:1
           description: DIMM DDR2 Synchronous 667 MHz (1.5 ns) [empty]
      *-bank:2
           description: DIMM DDR2 Synchronous 667 MHz (1.5 ns) [empty]
      *-bank:3
           description: DIMM DDR2 Synchronous 667 MHz (1.5 ns) [empty]

Answer 1

内存地址是"虚拟化的",程序使用的地址被转换为实际地址.这种转换使得有可能将您的程序看作连续内存的内容从当时的任何内容中分配出来.每个通用CPU都这样做.转换需要表查找,这需要内存访问.CPU有缓存,但长时间的虚拟地址很容易打破它的缓存,即"TLB"("转换后备缓冲区").因此,每隔4KB(Linux系统上的2MB就能找到你正在做的事情),CPU会停止寻找真正发送内存流量的地方.这些摊位可能需要相当多的时间.您可以尝试运行两个基准测试副本,TLB未命中的情况似乎是合理的,并且您将获得更接近您的额定容量的总带宽.

(编辑:嗯,你可能想要替换你#define的

size_t numBytes=atoi(argv[1]);
size_t numTransfers=atoi(argv[2]);

在主体......)

编辑:顺便说一句:我在我的盒子上看到的测试带宽(并在评论中报告)远远低于我的cpu的额定容量,它让我调查了我自己的系统.我的盒子构建器在这些插槽中放置了真正的垃圾记忆.我早已用一个知名品牌替换它们,报告的吞吐量增加了一倍以上,并且非常明显地改善了我的机器的性能.