user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xf8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xe8(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rcx
  lea    0xe8(%rsp),%rdx
  mov    %rcx,%rax
L30:
  movl   $0x0,(%rax)
  add    $0x18,%rax
  movl   $0x0,-0x14(%rax)
  movl   $0x0,-0x10(%rax)
  movl   $0x0,-0xc(%rax)
  movl   $0x0,-0x8(%rax)
  movl   $0x0,-0x4(%rax)
  cmp    %rdx,%rax
  jne    L30
  lea    0xbc(%rcx),%rax
  pxor   %xmm0,%xmm0
  lea    0x1c(%rsp),%r14
  xor    %ebp,%ebp
  mov    %rax,0x8(%rsp)
  lea    0xa4(%rcx),%rax
  lea    D27(%rip),%r15
  movq   $0x0,0x30(%rsp)
  movl   $0x0,0x38(%rsp)
  mov    %rax,(%rsp)
  movaps %xmm0,0x20(%rsp)
  nopl   0x0(%rax)
L71:
  mov    $0x1,%r12d
  cs nopw 0x0(%rax,%rax,1)
L101:
  mov    %r12d,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    %r14,%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  movslq 0x1c(%rsp),%rax
  cmp    $0x6,%eax
  ja     L101
  mov    0x20(%rsp,%rax,4),%edx
  cmp    $0x5,%edx
  jg     L101
  movslq %edx,%rcx
  lea    (%rax,%rax,2),%rsi
  add    $0x1,%edx
  mov    0x8(%rsp),%rbx
  lea    (%rcx,%rsi,2),%rcx
  mov    %edx,0x20(%rsp,%rax,4)
  mov    %r12d,0x40(%rsp,%rcx,4)
  xchg   %ax,%ax
L132:
  mov    $0x7c,%edi
  lea    -0xa8(%rbx),%r13
  call   <putchar@plt>
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L37:
  test   %eax,%eax
  je     L2
L56:
  add    $0x18,%r13
  cmp    %r13,%rbx
  je     L89
L58:
  mov    0x0(%r13),%eax
  cmp    $0x1,%eax
  je     L87
  cmp    $0x2,%eax
  jne    L37
  mov    $0x30,%edi
  add    $0x18,%r13
  call   <putchar@plt>
  cmp    %r13,%rbx
  jne    L58
L89:
  lea    D23(%rip),%rdi
  sub    $0x4,%rbx
  call   <puts@plt>
  cmp    (%rsp),%rbx
  jne    L132
  cmp    $0x2,%r12d
  je     L77
  mov    $0x2,%r12d
  jmp    L101
  nopl   (%rax)
L87:
  mov    $0x78,%edi
  call   <putchar@plt>
  jmp    L56
  nopl   0x0(%rax)
L2:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L56
  nopl   0x0(%rax)
L77:
  cmp    $0x29,%ebp
  jle    L71
  mov    0xe8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L104
  add    $0xf8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L104:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D27
type_infered: string
name: disassemble_data
arguments: label: D16
type_infered: string
name: disassemble_data
arguments: label: D23
type_infered: string

tool: D27: joueur %d 
Entrez une colonne (0-6) :

tool: D16: %d

tool: D23: |

assistant: ```c
int main() {
  int colonne;
  int plateau[7][6];
  int i, j;
  int compteur = 0;
  int joueur;
  int hauteurs[7];
  for (j = 0; j < 7; j++) {
    for (i = 0; i < 6; i++) {
      plateau[j][i] = 0;
    }
    hauteurs[j] = 0;
  }
  do {
    for (joueur = 1; joueur <= 2; joueur++) {
      do {
        printf("joueur %d \nEntrez une colonne (0-6) :", joueur);
        compteur++;
        scanf("%d", &colonne);
      } while (colonne < 0 || colonne > 6 || hauteurs[colonne] >= 6);
      plateau[colonne][hauteurs[colonne]] = joueur;
      hauteurs[colonne]++;
      for (i = 5; i >= 0; i--) {
        printf("|");
        for (j = 0; j < 7; j++) {
          switch (plateau[j][i]) {
          case 0:
            printf(" ");
            break;
          case 1:
            printf("x");
            break;
          case 2:
            printf("0");
            break;
          }
        }
        printf("|\n");
      }
    }
  } while (compteur < 42);
  return 0;
}
```