Re: [問題] c請益

[lungswu (宅爸爸)]

※ 引述《slind (你別對著螢幕傻笑,很蠢耶)》之銘言：
: void main(){
:   int b[3];
:   memset(b, 0, sizeof(b));
:   char *a=(char*)b;
:   int i;
:   for(i=0; i<9; i++){
:     a[i] = 0x5f;
:   }
:   printf("%x %x %x\n", b[0], b[1], b[2]);
: }
: 請問這段是怎麼運作的? 越詳細越好 感謝

我想它最詳細的運作方式是這樣的，再講下去要查Micro Instruction了

void main(){
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
   4:   48 83 ec 20             sub    $0x20,%rsp

  int b[3];
  memset(b, 0, sizeof(b));
   8:   48 8d 45 e0             lea    -0x20(%rbp),%rax
   c:   ba 0c 00 00 00          mov    $0xc,%edx
  11:   be 00 00 00 00          mov    $0x0,%esi
  16:   48 89 c7                mov    %rax,%rdi
  19:   e8 00 00 00 00          callq  1e <main+0x1e>

  char *a=(char*)b;
  1e:   48 8d 45 e0             lea    -0x20(%rbp),%rax
  22:   48 89 45 f0             mov    %rax,-0x10(%rbp)
  19:   e8 00 00 00 00          callq  1e <main+0x1e>

  char *a=(char*)b;
  1e:   48 8d 45 e0             lea    -0x20(%rbp),%rax
  22:   48 89 45 f0             mov    %rax,-0x10(%rbp)
  int i;
  for(i=0; i<9; i++){
  26:   c7 45 fc 00 00 00 00    movl   $0x0,-0x4(%rbp)
  2d:   eb 10                   jmp    3f <main+0x3f>
    a[i] = 0x5f;
  2f:   8b 45 fc                mov    -0x4(%rbp),%eax
  32:   48 98                   cltq
  34:   48 03 45 f0             add    -0x10(%rbp),%rax
  38:   c6 00 5f                movb   $0x5f,(%rax)
  int b[3];
  memset(b, 0, sizeof(b));

  char *a=(char*)b;
  int i;
  for(i=0; i<9; i++){
  3b:   83 45 fc 01             addl   $0x1,-0x4(%rbp)
  3f:   83 7d fc 08             cmpl   $0x8,-0x4(%rbp)
  43:   7e ea                   jle    2f <main+0x2f>
    a[i] = 0x5f;
  }

  printf("%x %x %x\n", b[0], b[1], b[2]);
  45:   8b 4d e8                mov    -0x18(%rbp),%ecx
  48:   8b 55 e4                mov    -0x1c(%rbp),%edx
  4b:   8b 45 e0                mov    -0x20(%rbp),%eax
  4e:   89 c6                   mov    %eax,%esi
  50:   bf 00 00 00 00          mov    $0x0,%edi
  55:   b8 00 00 00 00          mov    $0x0,%eax
  5a:   e8 00 00 00 00          callq  5f <main+0x5f>
}
  5f:   c9                      leaveq
  60:   c3                      retq

--
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◆ From: 134.122.249.1
→ BombCat:再再講下去要查NPN、PNP電晶體了
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推 diabloevagto:詳細推
推 stupid0319:64位元版本
※ 編輯: lungswu         來自: 134.122.249.1        (09/06 19:37)
推 VictorTom:push....XD
→ yzpdal:一目了然真清楚 謝謝解釋