c - 为什么 -O1 比 -O2 快

Question

我写了一个这样的C代码:

#include <stdio.h>
#define N 19

int main(void){
    int a[N];
    int ans = 0;

    for(int i = 0; i < N; ++i){
        a[i] = 0;
    }
    for(;;){
        int i;
        ++ans;
        for(i = N - 1; a[i] == 2; --i){
            if(i == 0){
                printf("%d\n", ans);
                return 0;
            }else{
                a[i] = 0;
            }
        }
        ++a[i];
    }
}

这会计算从 0 到 2 选择 N (= 19 ) 个数字的方式，并打印方式的数量 (= 3^19 = 1,162,261,467)。
我用 gcc 编译了这段代码。 -O1 比 -O2 快。为什么 -O2 优化比 -O1 差？

CPU:Intel(R) Core(TM) i7-8565U，x86_64

操作系统:Arch Linux (5.9.1-arch1-1)

编译器:gcc (GCC) 10.2.0

编辑:
使用 -S 选项运行 gcc 生成以下汇编代码:
-O1

    .file   "a.c"
    .text
    .section    .rodata.str1.1,"aMS",@progbits,1
.LC0:
    .string "%d\n"
    .text
    .globl  main
    .type   main, @function
main:
.LFB11:
    .cfi_startproc
    subq    $104, %rsp
    .cfi_def_cfa_offset 112
    movq    %fs:40, %rax
    movq    %rax, 88(%rsp)
    xorl    %eax, %eax
    movq    %rsp, %rax
    leaq    76(%rsp), %rdx
.L2:
    movl    $0, (%rax)
    addq    $4, %rax
    cmpq    %rdx, %rax
    jne .L2
    movl    $0, %esi
    jmp .L7
.L4:
    movslq  %edx, %rdx
    addl    $1, %ecx
    movl    %ecx, (%rsp,%rdx,4)
.L7:
    addl    $1, %esi
    movl    72(%rsp), %ecx
    leaq    68(%rsp), %rax
    movl    $18, %edx
    cmpl    $2, %ecx
    jne .L4
.L5:
    movl    $0, 4(%rax)
    subl    $1, %edx
    movl    (%rax), %ecx
    cmpl    $2, %ecx
    jne .L4
    subq    $4, %rax
    testl   %edx, %edx
    jne .L5
    leaq    .LC0(%rip), %rdi
    movl    $0, %eax
    call    printf@PLT
    movq    88(%rsp), %rax
    subq    %fs:40, %rax
    jne .L14
    movl    $0, %eax
    addq    $104, %rsp
    .cfi_remember_state
    .cfi_def_cfa_offset 8
    ret
.L14:
    .cfi_restore_state
    call    __stack_chk_fail@PLT
    .cfi_endproc
.LFE11:
    .size   main, .-main
    .ident  "GCC: (GNU) 10.2.0"
    .section    .note.GNU-stack,"",@progbits

-O2

    .file   "a.c"
    .text
    .section    .rodata.str1.1,"aMS",@progbits,1
.LC0:
    .string "%d\n"
    .section    .text.startup,"ax",@progbits
    .p2align 4
    .globl  main
    .type   main, @function
main:
.LFB11:
    .cfi_startproc
    subq    $104, %rsp
    .cfi_def_cfa_offset 112
    movl    $9, %ecx
    xorl    %esi, %esi
    movq    %fs:40, %rax
    movq    %rax, 88(%rsp)
    xorl    %eax, %eax
    movq    %rsp, %rdx
    movq    %rdx, %rdi
    rep stosq
    movl    $0, (%rdi)
    leaq    68(%rsp), %rdi
.L6:
    movl    72(%rsp), %ecx
    addl    $1, %esi
    movq    %rdi, %rax
    movl    $18, %edx
    cmpl    $2, %ecx
    je  .L4
    jmp .L3
    .p2align 4,,10
    .p2align 3
.L5:
    subq    $4, %rax
    testl   %edx, %edx
    je  .L14
.L4:
    movl    (%rax), %ecx
    movl    $0, 4(%rax)
    subl    $1, %edx
    cmpl    $2, %ecx
    je  .L5
.L3:
    movslq  %edx, %rdx
    addl    $1, %ecx
    movl    %ecx, (%rsp,%rdx,4)
    jmp .L6
    .p2align 4,,10
    .p2align 3
.L14:
    xorl    %eax, %eax
    leaq    .LC0(%rip), %rdi
    call    printf@PLT
    movq    88(%rsp), %rax
    subq    %fs:40, %rax
    jne .L15
    xorl    %eax, %eax
    addq    $104, %rsp
    .cfi_remember_state
    .cfi_def_cfa_offset 8
    ret
.L15:
    .cfi_restore_state
    call    __stack_chk_fail@PLT
    .cfi_endproc
.LFE11:
    .size   main, .-main
    .ident  "GCC: (GNU) 10.2.0"
    .section    .note.GNU-stack,"",@progbits

基准是:

$ gcc a.c -O1
$ time ./a.out
1162261467

real    0m0.895s
user    0m0.894s
sys 0m0.000s
$ time ./a.out
1162261467

real    0m0.912s
user    0m0.911s
sys 0m0.000s
$ time ./a.out
1162261467

real    0m0.925s
user    0m0.924s
sys 0m0.001s
$ gcc a.c -O2
$ time ./a.out
1162261467

real    0m1.570s
user    0m1.568s
sys 0m0.000s
$ time ./a.out
1162261467

real    0m1.567s
user    0m1.562s
sys 0m0.004s
$ time ./a.out
1162261467

real    0m1.576s
user    0m1.568s
sys 0m0.001s
$ gcc a.c -O3
$ time ./a.out
1162261467

real    0m1.613s
user    0m1.612s
sys 0m0.000s
$ time ./a.out
1162261467

real    0m1.608s
user    0m1.599s
sys 0m0.003s
$ time ./a.out
1162261467

real    0m1.628s
user    0m1.628s
sys 0m0.000s
$ gcc a.c -Ofast
$ time ./a.out
1162261467

real    0m1.571s
user    0m1.570s
sys 0m0.001s
$ time ./a.out
1162261467

real    0m1.604s
user    0m1.595s
sys 0m0.004s
$ time ./a.out
1162261467

real    0m1.616s
user    0m1.613s
sys 0m0.000s
$ gcc a.c -O0
$ time ./a.out
1162261467

real    0m2.457s
user    0m2.456s
sys 0m0.001s
$ time ./a.out
1162261467

real    0m2.526s
user    0m2.525s
sys 0m0.000s
$ time ./a.out
1162261467

real    0m2.565s
user    0m2.565s
sys 0m0.000s

编辑:
我编辑了这样的代码:

#include <stdio.h>
#define N 19

volatile int answer;

int main(void){
    int a[N];
    int ans = 0;

    for(int i = 0; i < N; ++i){
        a[i] = 0;
    }
    for(;;){
        int i;
        ++ans;
        for(i = N - 1; a[i] == 2; --i){
            if(i == 0){
                answer = ans;
                return 0;
            }else{
                a[i] = 0;
            }
        }
        ++a[i];
    }
}

并再次测量:

$ gcc a.c -O1
$ time ./a.out

real    0m0.924s
user    0m0.924s
sys 0m0.000s
$ time ./a.out

real    0m0.950s
user    0m0.949s
sys 0m0.000s
$ time ./a.out

real    0m0.993s
user    0m0.989s
sys 0m0.004s
$ gcc a.c -O2
$ time ./a.out

real    0m1.637s
user    0m1.636s
sys 0m0.000s
$ time ./a.out

real    0m1.661s
user    0m1.656s
sys 0m0.004s
$ time ./a.out

real    0m1.656s
user    0m1.654s
sys 0m0.001s

编辑:
我在 [[likely]] 之后添加了 for(;;) 属性:

#include <stdio.h>
#define N 19

int main(void){
    int a[N];
    int ans = 0;

    for(int i = 0; i < N; ++i){
        a[i] = 0;
    }
    for(;;) [[likely]] {
        int i;
        ++ans;
        for(i = N - 1; a[i] == 2; --i){
            if(i == 0){
                printf("%d\n", ans);
                return 0;
            }else{
                a[i] = 0;
            }
        }
        ++a[i];
    }
}

然后，基准测试的结果发生了变化:

$ g++ a.cpp -O1
$ for i in {1..5}; do time ./a.out; done
1162261467
./a.out  0.65s user 0.00s system 99% cpu 0.653 total
1162261467
./a.out  0.65s user 0.00s system 99% cpu 0.657 total
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.656 total
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.665 total
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.660 total
$ g++ a.cpp -O2
$ for i in {1..5}; do time ./a.out; done
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.661 total
1162261467
./a.out  0.65s user 0.00s system 99% cpu 0.648 total
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.659 total
1162261467
./a.out  0.65s user 0.00s system 99% cpu 0.654 total
1162261467
./a.out  0.66s user 0.00s system 99% cpu 0.657 total

-O2 和 -O1 一样快!谢谢@Acorn。

Answer 1

Why -O2 optimization was worse than -O1?

在大多数情况下，更高的优化级别应该会给你更好的性能。不过，您可能会发现像这样的异常(exception)。特别是在像这样的微基准测试中。
您的程序使用的代码和数据内存非常小，缓存和内存访问不太可能成为问题。但是，它是分支密集型的，这意味着它将归结为静态和动态分支预测。
如果您的编译器出错，就像在这种情况下一样，您可以尝试通过可能/不太可能的提示或分析程序为其提供更多信息。