Question

使用ocount对代码进行分析会显示更多循环，其中惩罚为惩罚，而较小的循环则为惩罚。我试图理解为什么惩罚标志开启时会有更多的惩罚？

uint16_t arr[1010];
uint32_t r[500];


void func()
{     
            uint32_t i = 0;


            for (i = 0; i < 1000; i+=2)
            {
                arr[i] = i;
                arr[i+1] = i+10;
        #ifdef PENALTY_ON

               r[i/2] = *(uint32_t *)((uint16_t *)&arr[i+1]);
        #endif
            }
        #ifndef PENALTY_ON
            for (i = 0; i < 1000; i+=2)
            {
                r[i/2] = *(uint32_t *)((uint16_t *)&arr[i+1]);
            }
        #endif
 }

Answer 1

使用-O3

在32位计算机上使用gcc进行编译

使用PENALTY_ON

00000000 <func>:
0:  31 c0                   xor    %eax,%eax
2:  8d b6 00 00 00 00       lea    0x0(%esi),%esi
8:  8d 50 0a                lea    0xa(%eax),%edx
b:  66 89 94 00 02 00 00    mov    %dx,0x2(%eax,%eax,1)
12: 00 
13: 8b 8c 00 02 00 00 00    mov    0x2(%eax,%eax,1),%ecx
1a: 89 c2                   mov    %eax,%edx
1c: 66 89 84 00 00 00 00    mov    %ax,0x0(%eax,%eax,1)
23: 00 
24: 83 c0 02                add    $0x2,%eax
27: d1 ea                   shr    %edx
29: 3d e8 03 00 00          cmp    $0x3e8,%eax
2e: 89 0c 95 00 00 00 00    mov    %ecx,0x0(,%edx,4)
35: 75 d1                   jne    8 <func+0x8>
37: f3 c3                   repz ret

没有PENALTY_ON

00000000 <func>:
0:  31 c0                   xor    %eax,%eax
2:  8d b6 00 00 00 00       lea    0x0(%esi),%esi
8:  8d 50 0a                lea    0xa(%eax),%edx
b:  66 89 84 00 00 00 00    mov    %ax,0x0(%eax,%eax,1)
12: 00 
13: 66 89 94 00 02 00 00    mov    %dx,0x2(%eax,%eax,1)
1a: 00 
1b: 83 c0 02                add    $0x2,%eax
1e: 3d e8 03 00 00          cmp    $0x3e8,%eax
23: 75 e3                   jne    8 <func+0x8>
25: 66 31 c0                xor    %ax,%ax
28: 8b 8c 00 02 00 00 00    mov    0x2(%eax,%eax,1),%ecx
2f: 89 c2                   mov    %eax,%edx
31: 83 c0 02                add    $0x2,%eax
34: d1 ea                   shr    %edx
36: 3d e8 03 00 00          cmp    $0x3e8,%eax
3b: 89 0c 95 00 00 00 00    mov    %ecx,0x0(,%edx,4)
42: 75 e4                   jne    28 <func+0x28>
44: f3 c3                   repz ret

我认为原因是PENALTY_ON

发生了写后读取停顿

b:  66 89 94 00 02 00 00    mov    %dx,0x2(%eax,%eax,1)
12: 00 
13: 8b 8c 00 02 00 00 00    mov    0x2(%eax,%eax,1),%ecx

内存大小加载和存储惩罚分析？

1 个答案: