user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_maze>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebp
  mov    $0x0,%ebx
  lea    D47(%rip),%r13
  lea    D32(%rip),%r14
  jmp    L99
L70:
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L133:
  add    $0x1,%ebx
  add    $0x1,%ebp
L99:
  cmp    $0x9,%ebx
  jle    L70
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x1d,%ebp
  jle    L133
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1e,%r12
  mov    $0x0,%r14d
  lea    D8(%rip),%rbp
  jmp    L155
L12:
  mov    %r14d,%esi
  lea    D79(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L164
  lea    D75(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L129:
  add    $0x1,%rbx
  cmp    %r12,%rbx
  je     L160
L54:
  movzbl (%rbx),%eax
  sub    $0x20,%eax
  cmp    $0x18,%al
  ja     L129
  movzbl %al,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  lea    D105(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  lea    D57(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L129
L160:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14d
  add    $0x1e,%r12
  cmp    $0x17,%r14d
  je     L18
L155:
  cmp    $0x9,%r14d
  jle    L12
  mov    %r14d,%esi
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L164:
  lea    -0x1e(%r12),%rbx
  lea    D18(%rip),%r13
  lea    D62(%rip),%r15
  jmp    L54
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D18
type_guessed: string
name: read_data
arguments: data_tag: D47
type_guessed: string
name: read_data
arguments: data_tag: D32
type_guessed: string
name: read_data
arguments: data_tag: D79
type_guessed: string
name: read_data
arguments: data_tag: D75
type_guessed: string
name: read_data
arguments: data_tag: D41
type_guessed: string
name: read_data
arguments: data_tag: D105
type_guessed: string
name: read_data
arguments: data_tag: D57
type_guessed: string
name: read_data
arguments: data_tag: D62
type_guessed: string

tool: D18:    

tool: D47: %d 

tool: D32:  %d 

tool: D79: %d  

tool: D75: [X]

tool: D41:  G 

tool: D105:  P 

tool: D57:  R 

tool: D62:  T 

assistant: ```c
void print_maze(char maze[23][30]) {
  int i, j, k;
  printf("   ");
  for (i = 0; i < 30; i++) {
    if (i < 10) {
      printf(" %d ", i);
    } else {
      printf("%d ", i);
    }
  }
  printf("\n");
  k = 0;
  for (i = 0; i < 23; i++) {
    if (k < 10) {
      printf("%d  ", k++);
    } else {
      printf("%d ", k++);
    }
    for (j = 0; j < 30; j++) {
      switch (maze[i][j]) {
      case '0':
        printf("[X]");
        break;
      case '7':
        printf("   ");
        break;
      case '6':
        printf("   ");
        break;
      case '3':
        printf(" G ");
        break;
      case '5':
        printf(" P ");
        break;
      case '4':
        printf(" R ");
        break;
      case '8':
        printf(" T ");
        break;
      case ' ':
        printf("   ");
        break;
      }
    }
    printf("\n");
  }
}
```