当我启用 C11 时，我在一个小的 C 片段中发现了一个有趣的性能回归：
#include <vector> struct Item { int a; int b; }; int main() { const std::size_t num_items = 10000000; std::vector<Item> 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 (prerelease) 和 C03 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%] <not supported> stalled-cycles-frontend <not supported> 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，启用 C++11 后，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%] <not supported> stalled-cycles-frontend <not supported> 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 STL 通过启用 C++11 变得更快，尤其是。感谢移动语义。
编辑： 正如建议的那样，使用container.emplace_back();性能与 C03 版本相当。C03 版本如何实现同样的效果push_back？
container.emplace_back();
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%] <not supported> stalled-cycles-frontend <not supported> 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% )
我可以使用您在帖子中写的那些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：
-flto
（我正在编译你的原始代码，用container.push_back(Item());）
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 C11 模式下，生成的代码比 C98 模式下 明显更混乱，并且在 C++11 模式下， 内联函数 void std::vector<Item,std::allocator<Item>>::_M_emplace_back_aux<Item>(Item&&) 失败 ，默认为inline-limit.
g++ -std=c++11 -O3 -S regr.cpp
void std::vector<Item,std::allocator<Item>>::_M_emplace_back_aux<Item>(Item&&)
inline-limit
这种失败的内联具有多米诺骨牌效应。 不是因为这个函数正在被调用（它甚至没有被调用！），而是因为我们必须做好准备： 如果 它被调用，函数参数（Item.a和Item.b）必须已经在正确的位置。这会导致代码非常混乱。
Item.a
Item.b
这是内联成功 的情况下生成的代码的相关部分：
.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 #,
这是一个漂亮而紧凑的 for 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_ #
这段代码很杂乱，循环中发生的事情比前一种情况多得多。在函数之前call（显示的最后一行），参数必须适当放置：
call
leaq 16(%rsp), %rsi #, leaq 32(%rsp), %rdi #, call _ZNSt6vectorI4ItemSaIS0_EE19_M_emplace_back_auxIIS0_EEEvDpOT_ #
即使这从未真正执行过，循环也会安排之前的事情：
movl $0, 16(%rsp) #, D.13379.a movl $0, 20(%rsp) #, D.13379.b
这会导致代码混乱。 如果因为内联成功而没有函数call，我们在循环中只有 2 个移动指令，并且%rsp（堆栈指针）没有混乱。但是，如果内联失败，我们会得到 6 步，我们会在%rsp.
%rsp
只是为了证实我的理论（注意-finline-limit），都在 C++11 模式下：
-finline-limit
$ 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% )
事实上，如果我们要求编译器稍微努力地内联该函数，性能上的差异就会消失。
那么这个故事有什么意义呢？失败的内联会花费你很多，你应该充分利用编译器的能力： 我只能推荐链接时间优化。 它显着提升了我的程序的性能（高达 2.5 倍），我需要做的就是通过-flto标志。这是一笔不错的交易！;)
但是，我不建议使用 inline 关键字丢弃您的代码。让编译器决定要做什么。（无论如何，优化器都可以将 inline 关键字视为空格。）
好问题，+1！
