为什么在同一个程序中相同C循环的相同副本需要显著但一致不同的时间来执行?
我希望我已经将我的问题简化为一个简单且可重复的测试用例。源(即)包含相同的简单循环的10个副本。每个循环的形式如下:为什么在同一个程序中相同C循环的相同副本需要显著但一致不同的时间来执行?,c,loops,assembly,intel,memory-alignment,C,Loops,Assembly,Intel,Memory Alignment,我希望我已经将我的问题简化为一个简单且可重复的测试用例。源(即)包含相同的简单循环的10个副本。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;
#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从内存读取和写入值。每个循环使用“-falign loops=64”与64个字节对齐,并生成相同的程序集,但与全局循环的相对偏移除外:
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>
我还在源代码中使用属性((noinline))使程序集更清晰,但观察问题时不需要这样做。我发现使用shell循环的函数速度最快,速度最慢:
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 ${n};
done
它产生的结果在每次运行中大约一致于1%,最快和最慢函数的确切数量根据确切的二进制布局而变化:
same-function 0:
2.08 seconds
2.04 seconds
2.06 seconds
same-function 1:
2.12 seconds
2.12 seconds
2.12 seconds
same-function 2:
2.10 seconds
2.14 seconds
2.11 seconds
same-function 3:
2.04 seconds
2.04 seconds
2.05 seconds
same-function 4:
2.05 seconds
2.00 seconds
2.03 seconds
same-function 5:
2.07 seconds
2.07 seconds
1.98 seconds
same-function 6:
1.83 seconds
1.83 seconds
1.83 seconds
same-function 7:
1.95 seconds
1.98 seconds
1.95 seconds
same-function 8:
1.86 seconds
1.88 seconds
1.86 seconds
same-function 9:
2.04 seconds
2.04 seconds
2.02 seconds
在本例中,我们看到loop2()是执行速度最慢的一个,而loop6()是执行速度最快的一个,两者之间的差异刚刚超过10%。我们通过使用不同的方法反复测试这两种情况来再次确认这一点:
nate@haswell$ 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 $N 2>&1 | grep GHz; done
6,234,770,361 cycles # 3.391 GHz
6,199,096,296 cycles # 3.391 GHz
6,213,348,126 cycles # 3.391 GHz
6,217,971,263 cycles # 3.391 GHz
6,224,779,686 cycles # 3.391 GHz
6,194,117,897 cycles # 3.391 GHz
6,225,259,274 cycles # 3.391 GHz
6,244,391,509 cycles # 3.391 GHz
6,189,972,381 cycles # 3.391 GHz
6,205,556,306 cycles # 3.391 GHz
考虑到这一点,我们重新阅读了每一本英特尔体系结构手册中的每一个单词,筛选了整个网络上提到“计算机”或“编程”的每一页,并在山顶上独自冥想了6年。由于无法获得任何形式的启蒙,我们来到文明世界,剃胡子,洗澡,并询问科学专家:
这里可能发生什么事
编辑:在本杰明的帮助下(见下面他的答案),我想出了一个更为有效的方法。这是一个独立的20线组装。从使用SUB更改为SBB会导致15%的性能差异,即使结果保持不变且执行的指令数相同。解释?我想我越来越接近了
; Minimal example, see also http://stackoverflow.com/q/26266953/3766665
; To build (Linux):
; nasm -felf64 func.asm
; ld func.o
; Then run:
; perf stat -r10 ./a.out
; On Haswell and Sandy Bridge, observed runtime varies
; ~15% depending on whether sub or sbb is used in the loop
section .text
global _start
_start:
push qword 0h ; put counter variable on stack
jmp loop ; jump to function
align 64 ; function alignment.
loop:
mov rcx, 1000000000
align 64 ; loop alignment.
l:
mov rax, [rsp]
add rax, 1h
mov [rsp], rax
; sbb rcx, 1h ; which is faster: sbb or sub?
sub rcx, 1h ; switch, time it, and find out
jne l ; (rot13 spoiler: foo vf snfgre ol 15%)
fin: ; If that was too easy, explain why.
mov eax, 60
xor edi, edi ; End of program. Exit with code 0
syscall
几年前,我会告诉你,当CPU到达任何一个循环时,检查CPU内部状态的任何差异;众所周知,这对无序预测过程(或类似过程)的能力有着深远的影响。例如,同一个循环的性能可能会发生高达15-20%的变化,这取决于CPU在进入循环之前所做的事情,而仅仅从两个不同的点跳转就足以改变执行速度 对你来说,这很难测试。您所要做的就是更改IF块中指令的顺序;例如,更换以下部件:
switch (firstLetter) {
case '0': loop0(); break;
case '1': loop1(); break;
case '2': loop2(); break;
case '3': loop3(); break;
case '4': loop4(); break;
case '5': loop5(); break;
case '6': loop6(); break;
case '7': loop7(); break;
case '8': loop8(); break;
case '9': loop9(); break;
default: goto die_usage;
}
与:
或者任意顺序。当然,您应该检查生成的汇编代码,以确保编译器没有对指令的顺序进行重新排序
此外,由于循环位于单个函数中;您还应该确保这些函数本身在64字节边界上对齐。查看完整的perf stat输出,您会发现变化的不是指令数,而是暂停的周期数 查看拆解,我发现了两件事:
(序号7);执行统计-e循环-r3./相同功能$f 2>&1;完成| grep循环
6070933420次循环(+-0.11%)
6052771142次循环(+-0.06%)
6099676333次循环(+-0.07%)
6092962697次循环(+-0.16%)
6151861993周期(+-0.69%)
6074323033次循环(+-0.36%)
6174434653循环(+-0.65%)
不过,我对你找到的摊位的性质不太了解
编辑:
我将计数器设置为每个函数中的易失性成员,在我的I7-3537U上测试了不同的编译,发现“-falign loops=64”实际上是最慢的:
$ 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% )
有关对齐循环与对齐功能的更多详细信息:
$ 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% )
及
但是没有找到罪犯。尽管如此,我还是觉得把它记在这里可能会对你有所帮助
编辑2:
这是我对相同功能的补丁。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:同样的事情还有更简单的例子:
; Minimal example, see also http://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
这里也有同样的效果。有趣
干杯,
Benjamin您如何解释后台进程中断您的程序?(我对Linux不太熟悉,但在Windows上,当尝试计时简单但长期运行的代码时,这是一个很大的问题。)当线程进入睡眠状态时,它会花费一个未知的时间间隔不运行,而计时器仍可能在滴答作响。这似乎不是处理器争用的问题。这可能解释了相同N的运行时间之间的差异,但不能解释不同N的运行时间之间一致的、统计上显著的差异。例如,
loop6
和loop1
的地址对齐方式是什么?它们都是32字节对齐的吗?他们有相同的总体定位吗?@NathanKurz你也禁用了涡轮增压器吗?您需要确保CPU不会动态缩放频率。另外,您是否尝试更改循环执行顺序?如果您执行进程10次(即,程序每次运行一次循环),则答案是缓存状态。如果你在里面重复这个循环
$ 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
$ 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
+}
; Minimal example, see also http://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