问
启用C++ 11时std :: vector performance regression
mobiledu2502912377 发布于 2023-02-05 22:16当我启用C++ 11时,我在一个小的C++片段中发现了一个有趣的性能回归:
#includestruct Item { int a; int b; }; int main() { const std::size_t num_items = 10000000; std::vector - container; container.reserve(num_items); for (std::size_t i = 0; i < num_items; ++i) { container.push_back(Item()); } return 0; }
使用g ++(GCC)4.8.2 20131219(预发行版)和C++ 03,我得到:
milian:/tmp$ g++ -O3 main.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
35.206824 task-clock # 0.988 CPUs utilized ( +- 1.23% )
4 context-switches # 0.116 K/sec ( +- 4.38% )
0 cpu-migrations # 0.006 K/sec ( +- 66.67% )
849 page-faults # 0.024 M/sec ( +- 6.02% )
95,693,808 cycles # 2.718 GHz ( +- 1.14% ) [49.72%]
stalled-cycles-frontend
stalled-cycles-backend
95,282,359 instructions # 1.00 insns per cycle ( +- 0.65% ) [75.27%]
30,104,021 branches # 855.062 M/sec ( +- 0.87% ) [77.46%]
6,038 branch-misses # 0.02% of all branches ( +- 25.73% ) [75.53%]
0.035648729 seconds time elapsed ( +- 1.22% )
另一方面,在启用C++ 11的情况下,性能会显着下降:
milian:/tmp$ g++ -std=c++11 -O3 main.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
86.485313 task-clock # 0.994 CPUs utilized ( +- 0.50% )
9 context-switches # 0.104 K/sec ( +- 1.66% )
2 cpu-migrations # 0.017 K/sec ( +- 26.76% )
798 page-faults # 0.009 M/sec ( +- 8.54% )
237,982,690 cycles # 2.752 GHz ( +- 0.41% ) [51.32%]
stalled-cycles-frontend
stalled-cycles-backend
135,730,319 instructions # 0.57 insns per cycle ( +- 0.32% ) [75.77%]
30,880,156 branches # 357.057 M/sec ( +- 0.25% ) [75.76%]
4,188 branch-misses # 0.01% of all branches ( +- 7.59% ) [74.08%]
0.087016724 seconds time elapsed ( +- 0.50% )
有人可以解释一下吗？到目前为止,我的经验是,通过启用C++ 11,特别是STL变得更快.感谢移动语义.
编辑:正如所建议的那样,使用container.emplace_back();性能与C++ 03版本相同.C++ 03版本如何实现相同的目标push_back？
milian:/tmp$ g++ -std=c++11 -O3 main.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
36.229348 task-clock # 0.988 CPUs utilized ( +- 0.81% )
4 context-switches # 0.116 K/sec ( +- 3.17% )
1 cpu-migrations # 0.017 K/sec ( +- 36.85% )
798 page-faults # 0.022 M/sec ( +- 8.54% )
94,488,818 cycles # 2.608 GHz ( +- 1.11% ) [50.44%]
stalled-cycles-frontend
stalled-cycles-backend
94,851,411 instructions # 1.00 insns per cycle ( +- 0.98% ) [75.22%]
30,468,562 branches # 840.991 M/sec ( +- 1.07% ) [76.71%]
2,723 branch-misses # 0.01% of all branches ( +- 9.84% ) [74.81%]
0.036678068 seconds time elapsed ( +- 0.80% )
Ali.. 246
我可以使用您在帖子中写下的选项在我的机器上重现您的结果.
但是,如果我还启用链接时间优化(我也将-flto标志传递给gcc 4.7.2),结果是相同的:
(我正在编译你的原始代码,container.push_back(Item());)
$ g++ -std=c++11 -O3 -flto regr.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
35.426793 task-clock # 0.986 CPUs utilized ( +- 1.75% )
4 context-switches # 0.116 K/sec ( +- 5.69% )
0 CPU-migrations # 0.006 K/sec ( +- 66.67% )
19,801 page-faults # 0.559 M/sec
99,028,466 cycles # 2.795 GHz ( +- 1.89% ) [77.53%]
50,721,061 stalled-cycles-frontend # 51.22% frontend cycles idle ( +- 3.74% ) [79.47%]
25,585,331 stalled-cycles-backend # 25.84% backend cycles idle ( +- 4.90% ) [73.07%]
141,947,224 instructions # 1.43 insns per cycle
# 0.36 stalled cycles per insn ( +- 0.52% ) [88.72%]
37,697,368 branches # 1064.092 M/sec ( +- 0.52% ) [88.75%]
26,700 branch-misses # 0.07% of all branches ( +- 3.91% ) [83.64%]
0.035943226 seconds time elapsed ( +- 1.79% )
$ g++ -std=c++98 -O3 -flto regr.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
35.510495 task-clock # 0.988 CPUs utilized ( +- 2.54% )
4 context-switches # 0.101 K/sec ( +- 7.41% )
0 CPU-migrations # 0.003 K/sec ( +-100.00% )
19,801 page-faults # 0.558 M/sec ( +- 0.00% )
98,463,570 cycles # 2.773 GHz ( +- 1.09% ) [77.71%]
50,079,978 stalled-cycles-frontend # 50.86% frontend cycles idle ( +- 2.20% ) [79.41%]
26,270,699 stalled-cycles-backend # 26.68% backend cycles idle ( +- 8.91% ) [74.43%]
141,427,211 instructions # 1.44 insns per cycle
# 0.35 stalled cycles per insn ( +- 0.23% ) [87.66%]
37,366,375 branches # 1052.263 M/sec ( +- 0.48% ) [88.61%]
26,621 branch-misses # 0.07% of all branches ( +- 5.28% ) [83.26%]
0.035953916 seconds time elapsed
至于原因,需要查看生成的汇编代码(g++ -std=c++11 -O3 -S regr.cpp).在C++ 11模式下,生成的代码比C++ 98模式更加混乱,并且在C++ 11模式下使用默认值内联函数
void std::vector
失败inline-limit.
此内联失败具有多米诺骨牌效应.不是因为这个函数被调用(它甚至没有调用!),而是因为我们必须做好准备:如果调用它,函数argments(Item.a和Item.b)必须已经在正确的位置.这会导致代码混乱.
以下是内联成功案例生成代码的相关部分:
.L42:
testq %rbx, %rbx # container$D13376$_M_impl$_M_finish
je .L3 #,
movl $0, (%rbx) #, container$D13376$_M_impl$_M_finish_136->a
movl $0, 4(%rbx) #, container$D13376$_M_impl$_M_finish_136->b
.L3:
addq $8, %rbx #, container$D13376$_M_impl$_M_finish
subq $1, %rbp #, ivtmp.106
je .L41 #,
.L14:
cmpq %rbx, %rdx # container$D13376$_M_impl$_M_finish, container$D13376$_M_impl$_M_end_of_storage
jne .L42 #,
这是一个很好的紧凑的循环.现在,让我们将其与失败的内联案例进行比较:
.L49:
testq %rax, %rax # D.15772
je .L26 #,
movq 16(%rsp), %rdx # D.13379, D.13379
movq %rdx, (%rax) # D.13379, *D.15772_60
.L26:
addq $8, %rax #, tmp75
subq $1, %rbx #, ivtmp.117
movq %rax, 40(%rsp) # tmp75, container.D.13376._M_impl._M_finish
je .L48 #,
.L28:
movq 40(%rsp), %rax # container.D.13376._M_impl._M_finish, D.15772
cmpq 48(%rsp), %rax # container.D.13376._M_impl._M_end_of_storage, D.15772
movl $0, 16(%rsp) #, D.13379.a
movl $0, 20(%rsp) #, D.13379.b
jne .L49 #,
leaq 16(%rsp), %rsi #,
leaq 32(%rsp), %rdi #,
call _ZNSt6vectorI4ItemSaIS0_EE19_M_emplace_back_auxIIS0_EEEvDpOT_ #
这段代码杂乱无章,循环中的内容比前一种情况要多得多.在函数call(显示最后一行)之前,必须正确放置参数:
leaq 16(%rsp), %rsi #, leaq 32(%rsp), %rdi #, call _ZNSt6vectorI4ItemSaIS0_EE19_M_emplace_back_auxIIS0_EEEvDpOT_ #
即使从未实际执行过,循环也会在之前排列:
movl $0, 16(%rsp) #, D.13379.a movl $0, 20(%rsp) #, D.13379.b
这会导致代码混乱.如果call由于内联成功没有函数,我们在循环中只有2个移动指令,并且没有%rsp(堆栈指针)的混乱.但是,如果内联失败,我们会得到6个动作,而且我们会乱用很多%rsp.
只是为了证实我的理论(注意-finline-limit),在C++ 11模式中:
$ g++ -std=c++11 -O3 -finline-limit=105 regr.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
84.739057 task-clock # 0.993 CPUs utilized ( +- 1.34% )
8 context-switches # 0.096 K/sec ( +- 2.22% )
1 CPU-migrations # 0.009 K/sec ( +- 64.01% )
19,801 page-faults # 0.234 M/sec
266,809,312 cycles # 3.149 GHz ( +- 0.58% ) [81.20%]
206,804,948 stalled-cycles-frontend # 77.51% frontend cycles idle ( +- 0.91% ) [81.25%]
129,078,683 stalled-cycles-backend # 48.38% backend cycles idle ( +- 1.37% ) [69.49%]
183,130,306 instructions # 0.69 insns per cycle
# 1.13 stalled cycles per insn ( +- 0.85% ) [85.35%]
38,759,720 branches # 457.401 M/sec ( +- 0.29% ) [85.43%]
24,527 branch-misses # 0.06% of all branches ( +- 2.66% ) [83.52%]
0.085359326 seconds time elapsed ( +- 1.31% )
$ g++ -std=c++11 -O3 -finline-limit=106 regr.cpp && perf stat -r 10 ./a.out
Performance counter stats for './a.out' (10 runs):
37.790325 task-clock # 0.990 CPUs utilized ( +- 2.06% )
4 context-switches # 0.098 K/sec ( +- 5.77% )
0 CPU-migrations # 0.011 K/sec ( +- 55.28% )
19,801 page-faults # 0.524 M/sec
104,699,973 cycles # 2.771 GHz ( +- 2.04% ) [78.91%]
58,023,151 stalled-cycles-frontend # 55.42% frontend cycles idle ( +- 4.03% ) [78.88%]
30,572,036 stalled-cycles-backend # 29.20% backend cycles idle ( +- 5.31% ) [71.40%]
140,669,773 instructions # 1.34 insns per cycle
# 0.41 stalled cycles per insn ( +- 1.40% ) [88.14%]
38,117,067 branches # 1008.646 M/sec ( +- 0.65% ) [89.38%]
27,519 branch-misses # 0.07% of all branches ( +- 4.01% ) [86.16%]
0.038187580 seconds time elapsed ( +- 2.05% )
实际上,如果我们要求编译器尝试更难以内联该函数,那么性能上的差异就会消失.
那么从这个故事中拿走了什么呢？内联失败可能会花费你很多,你应该充分利用编译器功能:我只能推荐链接时间优化.它为我的程序提供了显着的性能提升(高达2.5倍),我需要做的就是传递-flto旗帜.这是一个非常好的交易!;)
但是,我不建议使用inline关键字删除代码; 让编译器决定做什么.(无论如何,优化器都可以将内联关键字视为空格.)
好问题,+ 1!
1 个回答
-
我可以使用您在帖子中写下的选项在我的机器上重现您的结果.
但是,如果我还启用链接时间优化(我也将
-flto标志传递给gcc 4.7.2),结果是相同的:(我正在编译你的原始代码,
container.push_back(Item());)$ g++ -std=c++11 -O3 -flto regr.cpp && perf stat -r 10 ./a.out Performance counter stats for './a.out' (10 runs): 35.426793 task-clock # 0.986 CPUs utilized ( +- 1.75% ) 4 context-switches # 0.116 K/sec ( +- 5.69% ) 0 CPU-migrations # 0.006 K/sec ( +- 66.67% ) 19,801 page-faults # 0.559 M/sec 99,028,466 cycles # 2.795 GHz ( +- 1.89% ) [77.53%] 50,721,061 stalled-cycles-frontend # 51.22% frontend cycles idle ( +- 3.74% ) [79.47%] 25,585,331 stalled-cycles-backend # 25.84% backend cycles idle ( +- 4.90% ) [73.07%] 141,947,224 instructions # 1.43 insns per cycle # 0.36 stalled cycles per insn ( +- 0.52% ) [88.72%] 37,697,368 branches # 1064.092 M/sec ( +- 0.52% ) [88.75%] 26,700 branch-misses # 0.07% of all branches ( +- 3.91% ) [83.64%] 0.035943226 seconds time elapsed ( +- 1.79% ) $ g++ -std=c++98 -O3 -flto regr.cpp && perf stat -r 10 ./a.out Performance counter stats for './a.out' (10 runs): 35.510495 task-clock # 0.988 CPUs utilized ( +- 2.54% ) 4 context-switches # 0.101 K/sec ( +- 7.41% ) 0 CPU-migrations # 0.003 K/sec ( +-100.00% ) 19,801 page-faults # 0.558 M/sec ( +- 0.00% ) 98,463,570 cycles # 2.773 GHz ( +- 1.09% ) [77.71%] 50,079,978 stalled-cycles-frontend # 50.86% frontend cycles idle ( +- 2.20% ) [79.41%] 26,270,699 stalled-cycles-backend # 26.68% backend cycles idle ( +- 8.91% ) [74.43%] 141,427,211 instructions # 1.44 insns per cycle # 0.35 stalled cycles per insn ( +- 0.23% ) [87.66%] 37,366,375 branches # 1052.263 M/sec ( +- 0.48% ) [88.61%] 26,621 branch-misses # 0.07% of all branches ( +- 5.28% ) [83.26%] 0.035953916 seconds time elapsed至于原因,需要查看生成的汇编代码(
g++ -std=c++11 -O3 -S regr.cpp).在C++ 11模式下,生成的代码比C++ 98模式更加混乱,并且在C++ 11模式下使用默认值内联函数
void std::vector<Item,std::allocator<Item>>::_M_emplace_back_aux<Item>(Item&&)
失败inline-limit.此内联失败具有多米诺骨牌效应.不是因为这个函数被调用(它甚至没有调用!),而是因为我们必须做好准备:如果调用它,函数argments(
Item.a和Item.b)必须已经在正确的位置.这会导致代码混乱.以下是内联成功案例生成代码的相关部分:
.L42: testq %rbx, %rbx # container$D13376$_M_impl$_M_finish je .L3 #, movl $0, (%rbx) #, container$D13376$_M_impl$_M_finish_136->a movl $0, 4(%rbx) #, container$D13376$_M_impl$_M_finish_136->b .L3: addq $8, %rbx #, container$D13376$_M_impl$_M_finish subq $1, %rbp #, ivtmp.106 je .L41 #, .L14: cmpq %rbx, %rdx # container$D13376$_M_impl$_M_finish, container$D13376$_M_impl$_M_end_of_storage jne .L42 #,这是一个很好的紧凑的循环.现在,让我们将其与失败的内联案例进行比较:
.L49: testq %rax, %rax # D.15772 je .L26 #, movq 16(%rsp), %rdx # D.13379, D.13379 movq %rdx, (%rax) # D.13379, *D.15772_60 .L26: addq $8, %rax #, tmp75 subq $1, %rbx #, ivtmp.117 movq %rax, 40(%rsp) # tmp75, container.D.13376._M_impl._M_finish je .L48 #, .L28: movq 40(%rsp), %rax # container.D.13376._M_impl._M_finish, D.15772 cmpq 48(%rsp), %rax # container.D.13376._M_impl._M_end_of_storage, D.15772 movl $0, 16(%rsp) #, D.13379.a movl $0, 20(%rsp) #, D.13379.b jne .L49 #, leaq 16(%rsp), %rsi #, leaq 32(%rsp), %rdi #, call _ZNSt6vectorI4ItemSaIS0_EE19_M_emplace_back_auxIIS0_EEEvDpOT_ #这段代码杂乱无章,循环中的内容比前一种情况要多得多.在函数
call(显示最后一行)之前,必须正确放置参数:leaq 16(%rsp), %rsi #, leaq 32(%rsp), %rdi #, call _ZNSt6vectorI4ItemSaIS0_EE19_M_emplace_back_auxIIS0_EEEvDpOT_ #
即使从未实际执行过,循环也会在之前排列:
movl $0, 16(%rsp) #, D.13379.a movl $0, 20(%rsp) #, D.13379.b
这会导致代码混乱.如果
call由于内联成功没有函数,我们在循环中只有2个移动指令,并且没有%rsp(堆栈指针)的混乱.但是,如果内联失败,我们会得到6个动作,而且我们会乱用很多%rsp.只是为了证实我的理论(注意
-finline-limit),在C++ 11模式中:$ g++ -std=c++11 -O3 -finline-limit=105 regr.cpp && perf stat -r 10 ./a.out Performance counter stats for './a.out' (10 runs): 84.739057 task-clock # 0.993 CPUs utilized ( +- 1.34% ) 8 context-switches # 0.096 K/sec ( +- 2.22% ) 1 CPU-migrations # 0.009 K/sec ( +- 64.01% ) 19,801 page-faults # 0.234 M/sec 266,809,312 cycles # 3.149 GHz ( +- 0.58% ) [81.20%] 206,804,948 stalled-cycles-frontend # 77.51% frontend cycles idle ( +- 0.91% ) [81.25%] 129,078,683 stalled-cycles-backend # 48.38% backend cycles idle ( +- 1.37% ) [69.49%] 183,130,306 instructions # 0.69 insns per cycle # 1.13 stalled cycles per insn ( +- 0.85% ) [85.35%] 38,759,720 branches # 457.401 M/sec ( +- 0.29% ) [85.43%] 24,527 branch-misses # 0.06% of all branches ( +- 2.66% ) [83.52%] 0.085359326 seconds time elapsed ( +- 1.31% ) $ g++ -std=c++11 -O3 -finline-limit=106 regr.cpp && perf stat -r 10 ./a.out Performance counter stats for './a.out' (10 runs): 37.790325 task-clock # 0.990 CPUs utilized ( +- 2.06% ) 4 context-switches # 0.098 K/sec ( +- 5.77% ) 0 CPU-migrations # 0.011 K/sec ( +- 55.28% ) 19,801 page-faults # 0.524 M/sec 104,699,973 cycles # 2.771 GHz ( +- 2.04% ) [78.91%] 58,023,151 stalled-cycles-frontend # 55.42% frontend cycles idle ( +- 4.03% ) [78.88%] 30,572,036 stalled-cycles-backend # 29.20% backend cycles idle ( +- 5.31% ) [71.40%] 140,669,773 instructions # 1.34 insns per cycle # 0.41 stalled cycles per insn ( +- 1.40% ) [88.14%] 38,117,067 branches # 1008.646 M/sec ( +- 0.65% ) [89.38%] 27,519 branch-misses # 0.07% of all branches ( +- 4.01% ) [86.16%] 0.038187580 seconds time elapsed ( +- 2.05% )实际上,如果我们要求编译器尝试更难以内联该函数,那么性能上的差异就会消失.
那么从这个故事中拿走了什么呢？内联失败可能会花费你很多,你应该充分利用编译器功能:我只能推荐链接时间优化.它为我的程序提供了显着的性能提升(高达2.5倍),我需要做的就是传递
-flto旗帜.这是一个非常好的交易!;)但是,我不建议使用inline关键字删除代码; 让编译器决定做什么.(无论如何,优化器都可以将内联关键字视为空格.)
好问题,+ 1!
2023-02-05 22:32 回答
俊惠芸菁亚扬
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