为什么同一个程序中相同的C循环的相同副本会显着地但一贯的不同的时间执行?

我希望我将我的问题简化到可重复的测试用例。源(即here)包含10个相同简单循环的副本。每个循环的形式如下:

#define COUNT (1000 * 1000 * 1000)
volatile uint64_t counter = 0;

void loopN(void) {
  for (int j = COUNT; j != 0; j--) {
    uint64_t val = counter;
    val = val + 1;
    counter = val;
  }
  return;
}

变量的“volatile”很重要,因为它强制在每次迭代时读取和写入内存的值。每个循环使用’-falign-loops = 64’对齐到64个字节,并产生相同的程序集,除了相对于全局的偏移量:

   400880:       48 8b 15 c1 07 20 00    mov    0x2007c1(%rip),%rdx  # 601048 <counter>
   400887:       48 83 c2 01             add    $0x1,%rdx
   40088b:       83 e8 01                sub    $0x1,%eax
   40088e:       48 89 15 b3 07 20 00    mov    %rdx,0x2007b3(%rip)  # 601048 <counter>
   400895:       75 e9                   jne    400880 <loop8+0x20>

我在英特尔Haswell i7-4470上运行Linux 3.11。我正在使用GCC 4.8.1和命令行编译程序:

 gcc -std=gnu99 -O3 -falign-loops=64 -Wall -Wextra same-function.c -o same-function

我还在源代码中使用属性((noinline))来使程序集更清晰,但这并不是必须遵守的。我找到一个shell循环中最快和最慢的函数:

for n in 0 1 2 3 4 5 6 7 8 9; 
do echo same-function ${n}:; 
/usr/bin/time -f "%e seconds" same-function ${n}; 
/usr/bin/time -f "%e seconds" same-function ${n}; 
/usr/bin/time -f "%e seconds" same-function ${n}; 
done

它产生的结果与从运行到运行的约1%一致,最快和最慢的函数的确切数目根据准确的二进制布局而变化:

same-function 0:
2.08 seconds
2.04 seconds
2.06 seconds
same-function 1:
2.12 seconds
2.12 seconds
2.12 seconds
same-function 2:
2.10 seconds
2.14 seconds
2.11 seconds
same-function 3:
2.04 seconds
2.04 seconds
2.05 seconds
same-function 4:
2.05 seconds
2.00 seconds
2.03 seconds
same-function 5:
2.07 seconds
2.07 seconds
1.98 seconds
same-function 6:
1.83 seconds
1.83 seconds
1.83 seconds
same-function 7:
1.95 seconds
1.98 seconds
1.95 seconds
same-function 8:
1.86 seconds
1.88 seconds
1.86 seconds
same-function 9:
2.04 seconds
2.04 seconds
2.02 seconds

在这种情况下,我们看到,loop2()是执行速度最慢的一个,loop6()是最快的,其差异只有10%以上。我们通过不同的方法重复测试这两种情况来重新确认:

nate@haswell$ N=2; for i in {1..10}; do perf stat same-function $N 2>&1 | grep GHz; done
     7,180,104,866 cycles                    #    3.391 GHz
     7,169,930,711 cycles                    #    3.391 GHz
     7,150,190,394 cycles                    #    3.391 GHz
     7,188,959,096 cycles                    #    3.391 GHz
     7,177,272,608 cycles                    #    3.391 GHz
     7,093,246,955 cycles                    #    3.391 GHz
     7,210,636,865 cycles                    #    3.391 GHz
     7,239,838,211 cycles                    #    3.391 GHz
     7,172,716,779 cycles                    #    3.391 GHz
     7,223,252,964 cycles                    #    3.391 GHz

nate@haswell$ N=6; for i in {1..10}; do perf stat same-function $N 2>&1 | grep GHz; done
     6,234,770,361 cycles                    #    3.391 GHz
     6,199,096,296 cycles                    #    3.391 GHz
     6,213,348,126 cycles                    #    3.391 GHz
     6,217,971,263 cycles                    #    3.391 GHz
     6,224,779,686 cycles                    #    3.391 GHz
     6,194,117,897 cycles                    #    3.391 GHz
     6,225,259,274 cycles                    #    3.391 GHz
     6,244,391,509 cycles                    #    3.391 GHz
     6,189,972,381 cycles                    #    3.391 GHz
     6,205,556,306 cycles                    #    3.391 GHz

考虑到这一点,我们重新阅读了每个英特尔架构手册中的每个单词,筛选出提到“计算机”或“程序设计”一词的整个网页上的每个页面,并在山顶上独立沉思了6年。不能实现任何启发,我们来到文明,刮胡子,洗澡,问StackOverflow的专家:

这里可能发生什么?

编辑:随着本杰明的帮助(见下面的答案),我已经想出了更多的succinct test case.这是一个独立的20行装配。即使结果保持不变,执行相同数量的指令,从使用SUB更改为SBB也会导致性能差异为15%。说明?我想我越来越接近了。

; Minimal example, see also https://stackoverflow.com/q/26266953/3766665
; To build (Linux):
;   nasm -felf64 func.asm
;   ld func.o
; Then run:
;   perf stat -r10 ./a.out
; On Haswell and Sandy Bridge, observed runtime varies 
; ~15% depending on whether sub or sbb is used in the loop
section .text
global _start
_start:
  push qword 0h       ; put counter variable on stack
  jmp loop            ; jump to function
align 64              ; function alignment.
loop:
  mov rcx, 1000000000
align 64              ; loop alignment.
l:
  mov rax, [rsp]
  add rax, 1h
  mov [rsp], rax
; sbb rcx, 1h         ; which is faster: sbb or sub?
  sub rcx, 1h         ; switch, time it, and find out
  jne l               ; (rot13 spoiler: foo vf snfgre ol 15%)
fin:                  ; If that was too easy, explain why.
  mov eax, 60
  xor edi, edi        ; End of program. Exit with code 0
  syscall
最佳答案
查看完整的输出状态输出,您将看到不是指令数量不同而是停滞的周期数。

看着拆卸,我发现了两件事情:

>计数器变量的偏移量在函数之间变化。尽管如此,对每个功能进行反制都没有使行为消失。
>这些功能不会放在64字节的边界上,所以它们可以覆盖不同数量的高速缓存线。使用-falign-functions = 64编译确实使差异几乎完全消失。

在我的机器上进行上述更改的测试,然后得到:

以$(seq 7)为单位; do perf stat -e cycles -r3 ./same-function $ f 2>& 1;完成| grep循环
     6,070,933,420个周期( – 0.11%)
     6,052,771,142个周期( – 0.06%)
     6,099,676,333个周期( – 0.07%)
     6,092,962,697个周期( – 0.16%)
     6,151,861,993个周期( – 0.69%)
     6,074,323,033个周期( – 0.36%)
     6,174,434,653个周期(-0.65%)

不过,我不太了解你发现的摊位的性质。

编辑:
我已经在每个功能中摆脱了一个不稳定的成员,测试了我的I7-3537U上的不同编译,实际上发现`-falign-loops = 64’是最慢的:

$ gcc  -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function
$ gcc -falign-loops=64 -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function-l64
$ gcc -falign-functions=64 -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function-f64
$ for prog in same-function{,-l64,-f64}; do echo $prog; for f in $(seq 7); do perf stat -e cycles -r10 ./$prog $f 2>&1; done|grep cycl; done
same-function
     6,079,966,292      cycles                     ( +-  0.19% )
     7,419,053,569      cycles                     ( +-  0.07% )
     6,136,061,105      cycles                     ( +-  0.27% )
     7,282,434,896      cycles                     ( +-  0.74% )
     6,104,866,406      cycles                     ( +-  0.16% )
     7,342,985,942      cycles                     ( +-  0.52% )
     6,208,373,040      cycles                     ( +-  0.50% )
same-function-l64
     7,336,838,175      cycles                     ( +-  0.46% )
     7,358,913,923      cycles                     ( +-  0.52% )
     7,412,570,515      cycles                     ( +-  0.38% )
     7,435,048,756      cycles                     ( +-  0.10% )
     7,404,834,458      cycles                     ( +-  0.34% )
     7,291,095,582      cycles                     ( +-  0.99% )
     7,312,052,598      cycles                     ( +-  0.95% )
same-function-f64
     6,103,059,996      cycles                     ( +-  0.12% )
     6,116,601,533      cycles                     ( +-  0.29% )
     6,120,841,824      cycles                     ( +-  0.18% )
     6,114,278,098      cycles                     ( +-  0.09% )
     6,105,938,586      cycles                     ( +-  0.14% )
     6,101,672,717      cycles                     ( +-  0.19% )
     6,121,339,944      cycles                     ( +-  0.11% )

align-loops对对齐函数的更多细节:

$ for prog in same-function{-l64,-f64}; do sudo perf stat -d -r10 ./$prog 0; done

 Performance counter stats for './same-function-l64 0' (10 runs):

       2396.608194      task-clock:HG (msec)      #    1.001 CPUs utilized            ( +-  0.64% )
                56      context-switches:HG       #    0.024 K/sec                    ( +-  5.51% )
                 1      cpu-migrations:HG         #    0.000 K/sec                    ( +- 74.78% )
                46      page-faults:HG            #    0.019 K/sec                    ( +-  0.63% )
     7,331,450,530      cycles:HG                 #    3.059 GHz                      ( +-  0.51% ) [85.68%]
     5,332,248,218      stalled-cycles-frontend:HG #   72.73% frontend cycles idle     ( +-  0.71% ) [71.42%]
   <not supported>      stalled-cycles-backend:HG
     5,000,800,933      instructions:HG           #    0.68  insns per cycle
                                                  #    1.07  stalled cycles per insn  ( +-  0.04% ) [85.73%]
     1,000,446,303      branches:HG               #  417.443 M/sec                    ( +-  0.04% ) [85.75%]
             8,461      branch-misses:HG          #    0.00% of all branches          ( +-  6.05% ) [85.76%]
   <not supported>      L1-dcache-loads:HG
            45,593      L1-dcache-load-misses:HG  #    0.00% of all L1-dcache hits    ( +-  3.61% ) [85.77%]
             6,148      LLC-loads:HG              #    0.003 M/sec                    ( +-  8.80% ) [71.36%]
   <not supported>      LLC-load-misses:HG

       2.394456699 seconds time elapsed                                          ( +-  0.64% )


 Performance counter stats for './same-function-f64 0' (10 runs):

       1998.936383      task-clock:HG (msec)      #    1.001 CPUs utilized            ( +-  0.61% )
                60      context-switches:HG       #    0.030 K/sec                    ( +- 17.77% )
                 1      cpu-migrations:HG         #    0.001 K/sec                    ( +- 47.86% )
                46      page-faults:HG            #    0.023 K/sec                    ( +-  0.68% )
     6,107,877,836      cycles:HG                 #    3.056 GHz                      ( +-  0.34% ) [85.63%]
     4,112,602,649      stalled-cycles-frontend:HG #   67.33% frontend cycles idle     ( +-  0.52% ) [71.41%]
   <not supported>      stalled-cycles-backend:HG
     5,000,910,172      instructions:HG           #    0.82  insns per cycle
                                                  #    0.82  stalled cycles per insn  ( +-  0.01% ) [85.72%]
     1,000,423,026      branches:HG               #  500.478 M/sec                    ( +-  0.02% ) [85.77%]
            10,660      branch-misses:HG          #    0.00% of all branches          ( +- 13.23% ) [85.80%]
   <not supported>      L1-dcache-loads:HG
            47,492      L1-dcache-load-misses:HG  #    0.00% of all L1-dcache hits    ( +- 14.82% ) [85.80%]
            11,719      LLC-loads:HG              #    0.006 M/sec                    ( +- 42.44% ) [71.28%]
   <not supported>      LLC-load-misses:HG

       1.997319759 seconds time elapsed                                          ( +-  0.62% )

两个可执行文件都有非常低的指令/周期数,这可能是由于循环的简约性质和它所造成的内存压力,但我不知道为什么一个比另一个更糟。

我也尝试过一些事情

$ for prog in same-function{-l64,-f64}; do sudo perf stat -eL1-{d,i}cache-load-misses,L1-dcache-store-misses,cs,cycles,instructions -r10 ./$prog 0; done

$ sudo perf record -F25000 -e'{cycles:pp,stalled-cycles-frontend}' ./same-function-l64 0
[ perf record: Woken up 28 times to write data ]
[ perf record: Captured and wrote 6.771 MB perf.data (~295841 samples) ]
$ sudo perf report --group -Sloop0 -n --show-total-period --stdio
$ sudo perf annotate --group -sloop0  --stdio

但是找不到罪魁祸首没有任何成功。不过,我觉得这可能是有帮助的,无论如何,在这里记录下来…

编辑2:
这是我的patch to same-function.c:

$ git diff -u -U0
diff --git a/same-function.c b/same-function.c
index f78449e..78a5772 100644
--- a/same-function.c
+++ b/same-function.c
@@ -20 +20 @@ done
-volatile uint64_t counter = 0;
+//volatile uint64_t counter = 0;
@@ -22,0 +23 @@ COMPILER_NO_INLINE void loop0(void) {
+volatile uint64_t counter = 0;
@@ -31,0 +33 @@ COMPILER_NO_INLINE void loop1(void) {
+volatile uint64_t counter = 0;
@@ -40,0 +43 @@ COMPILER_NO_INLINE void loop2(void) {
+volatile uint64_t counter = 0;
@@ -49,0 +53 @@ COMPILER_NO_INLINE void loop3(void) {
+volatile uint64_t counter = 0;
@@ -58,0 +63 @@ COMPILER_NO_INLINE void loop4(void) {
+volatile uint64_t counter = 0;
@@ -67,0 +73 @@ COMPILER_NO_INLINE void loop5(void) {
+volatile uint64_t counter = 0;
@@ -76,0 +83 @@ COMPILER_NO_INLINE void loop6(void) {
+volatile uint64_t counter = 0;
@@ -85,0 +93 @@ COMPILER_NO_INLINE void loop7(void) {
+volatile uint64_t counter = 0;
@@ -94,0 +103 @@ COMPILER_NO_INLINE void loop8(void) {
+volatile uint64_t counter = 0;
@@ -103,0 +113 @@ COMPILER_NO_INLINE void loop9(void) {
+volatile uint64_t counter = 0;
@@ -135 +145 @@ int main(int argc, char** argv) {
-}
\ No newline at end of file
+}

编辑3:同样的事情甚至更少的例子:

; Minimal example, see also https://stackoverflow.com/q/26266953/3766665 
; To build (Linux):
;   nasm -felf64 func.asm
;   ld func.o
; Then run:
;   perf stat -r10 ./a.out
; Runtime varies ~10% depending on whether 
section .text
global _start
_start:
  push qword 0h       ; put counter variable on stack
  jmp loop            ; jump to function
;align 64             ; function alignment. Try commenting this
loop:
  mov rcx, 1000000000
;align 64             ; loop alignment. Try commenting this
l:
  mov rax, [rsp]
  add rax, 1h
  mov [rsp], rax
  sub rcx, 1h
  jne l
fin:                  ; End of program. Exit with code 0
  mov eax, 60
  xor edi, edi
  syscall

相同的效果在这里有趣。

干杯,
   本杰明

转载注明原文:为什么同一个程序中相同的C循环的相同副本会显着地但一贯的不同的时间执行? - 代码日志