user: 这段汇编代码用C语言表示是什么样的:

```asm
<print_maze>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  push   %r13
  lea    D106(%rip),%r13
  push   %r12
  lea    D67(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x18,%rsp
  call   <printf@plt>
  xor    %esi,%esi
  lea    0x1(%rsi),%ebx
  cmp    $0x9,%esi
  jbe    L152
  nopw   0x0(%rax,%rax,1)
L125:
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1e,%ebx
  je     L49
  mov    %ebx,%esi
L120:
  lea    0x1(%rsi),%ebx
  cmp    $0x9,%esi
  ja     L125
L152:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  jmp    L120
L49:
  mov    $0xa,%edi
  add    $0x1e,%rbp
  lea    D11(%rip),%rbx
  call   <putchar@plt>
  movl   $0x0,0xc(%rsp)
  nopl   0x0(%rax,%rax,1)
L104:
  mov    0xc(%rsp),%esi
  cmp    $0x9,%esi
  jbe    L154
  mov    0xc(%rsp),%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L161:
  lea    -0x1e(%rbp),%r15
  lea    D17(%rip),%r14
  lea    D102(%rip),%r13
  nopl   0x0(%rax)
L164:
  movzbl (%r15),%eax
  sub    $0x20,%eax
  cmp    $0x18,%al
  ja     L25
  movzbl %al,%eax
  movslq (%rbx,%rax,4),%rax
  add    %rbx,%rax
  notrack jmp *%rax
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L25:
  add    $0x1,%r15
  cmp    %r15,%rbp
  jne    L164
L103:
  mov    $0xa,%edi
  add    $0x1e,%rbp
  call   <putchar@plt>
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  cmp    $0x17,%eax
  jne    L104
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
  lea    D91(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L164
  jmp    L103
  nopl   0x0(%rax)
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L164
  jmp    L103
  nopl   0x0(%rax)
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L164
  jmp    L103
  nopl   0x0(%rax)
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L164
  jmp    L103
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L164
  jmp    L103
L154:
  lea    D103(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L161
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D106
type: string
name: extract_symbolic_data
arguments: data_label: D67
type: string
name: extract_symbolic_data
arguments: data_label: D7
type: string
name: extract_symbolic_data
arguments: data_label: D17
type: string
name: extract_symbolic_data
arguments: data_label: D102
type: string
name: extract_symbolic_data
arguments: data_label: D91
type: string
name: extract_symbolic_data
arguments: data_label: D3
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string
name: extract_symbolic_data
arguments: data_label: D103
type: string

tool: D106:  %d 

tool: D67:    

tool: D7: %d 

tool: D17:  R 

tool: D102:  P 

tool: D91:  T 

tool: D3:  G 

tool: D16: [X]

tool: D103: %d  

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");
  }
}
```