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    D20(%rip),%r13
  push   %r12
  lea    D72(%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    L31
  nopw   0x0(%rax,%rax,1)
L33:
  xor    %eax,%eax
  lea    D67(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1e,%ebx
  je     L173
  mov    %ebx,%esi
L63:
  lea    0x1(%rsi),%ebx
  cmp    $0x9,%esi
  ja     L33
L31:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  jmp    L63
L173:
  mov    $0xa,%edi
  add    $0x1e,%rbp
  lea    D5(%rip),%rbx
  call   <putchar@plt>
  movl   $0x0,0xc(%rsp)
  nopl   0x0(%rax,%rax,1)
L42:
  mov    0xc(%rsp),%esi
  cmp    $0x9,%esi
  jbe    L53
  lea    D67(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L52:
  lea    -0x1e(%rbp),%r15
  lea    D36(%rip),%r14
  lea    D35(%rip),%r13
  nopl   (%rax)
L93:
  movzbl (%r15),%eax
  sub    $0x20,%eax
  cmp    $0x18,%al
  ja     L55
  movzbl %al,%eax
  movslq (%rbx,%rax,4),%rax
  add    %rbx,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L55:
  add    $0x1,%r15
  cmp    %r15,%rbp
  jne    L93
L151:
  mov    $0xa,%edi
  add    $0x1e,%rbp
  call   <putchar@plt>
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  cmp    $0x17,%eax
  jne    L42
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
  lea    D77(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L93
  jmp    L151
  nopl   0x0(%rax)
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L93
  jmp    L151
  nopl   0x0(%rax)
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L93
  jmp    L151
  nopl   0x0(%rax)
  lea    D80(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L93
  jmp    L151
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jne    L93
  jmp    L151
L53:
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L52
```

tool_calls: name: parse_data
arguments: tag: D20
type: string
name: parse_data
arguments: tag: D72
type: string
name: parse_data
arguments: tag: D67
type: string
name: parse_data
arguments: tag: D36
type: string
name: parse_data
arguments: tag: D35
type: string
name: parse_data
arguments: tag: D77
type: string
name: parse_data
arguments: tag: D80
type: string
name: parse_data
arguments: tag: D62
type: string
name: parse_data
arguments: tag: D1
type: string

tool: D20:  %d 

tool: D72:    

tool: D67: %d 

tool: D36:  R 

tool: D35:  P 

tool: D77:  T 

tool: D80:  G 

tool: D62: [X]

tool: D1: %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");
  }
}
```