user: 请从给定的汇编指令推导出C源代码:

```asm
<libbf_naive_interpreter_int>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x60000(%rsp),%r11
L142:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L142
  sub    $0x18,%rsp
  mov    $0x20000,%edx
  xor    %esi,%esi
  xor    %r12d,%r12d
  lea    D5(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x60008(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rbx
  mov    %rsp,%rdi
  lea    0x20000(%rsp),%r13
  call   <memset@plt>
  xchg   %ax,%ax
L167:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L146
L29:
  sub    $0x2b,%eax
  cmp    $0x32,%al
  ja     L77
  movzbl %al,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  nopl   0x0(%rax)
  addl   $0x1,(%rsp,%r12,4)
  nopl   (%rax)
L77:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L29
L146:
  mov    0x60008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x60018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
  mov    -0x8(%r13),%rbx
  sub    $0x8,%r13
  jmp    L167
  nopw   0x0(%rax,%rax,1)
  mov    (%rsp,%r12,4),%eax
  test   %eax,%eax
  je     L46
  mov    %rbx,0x0(%r13)
  add    $0x1,%rbx
  add    $0x8,%r13
  jmp    L167
  nopl   (%rax)
  add    $0x1,%r12
  add    $0x1,%rbx
  jmp    L167
  nopl   (%rax)
  sub    $0x1,%r12
  add    $0x1,%rbx
  jmp    L167
  nopl   (%rax)
  mov    (%rsp,%r12,4),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  jmp    L167
  nopw   0x0(%rax,%rax,1)
  subl   $0x1,(%rsp,%r12,4)
  add    $0x1,%rbx
  jmp    L167
  cs nopw 0x0(%rax,%rax,1)
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <getchar@plt>
  mov    %eax,(%rsp,%r12,4)
  jmp    L167
  nopl   0x0(%rax)
L46:
  mov    $0x1,%ecx
  jmp    L98
  nop
L91:
  cmp    $0x5d,%al
  je     L117
L69:
  mov    %rdx,%rbx
L98:
  movzbl 0x1(%rbx),%eax
  lea    0x1(%rbx),%rdx
  cmp    $0x5b,%al
  jne    L91
  add    $0x1,%ecx
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L117:
  sub    $0x1,%ecx
  jne    L69
  add    $0x2,%rbx
  jmp    L167
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
type_guessed: string

tool: 

assistant: ```c
void libbf_naive_interpreter_int(const char *code) {
  char *code_ptr = (char *)code;
  unsigned int data[32768] = {0};
  long data_indice = 0;
  char *stack[32768];
  char **p_stack = stack;
  int level;
  while (1) {
    switch (*code_ptr) {
    case 0:
      return;
      break;
    case '+':
      data[data_indice]++;
      break;
    case '-':
      data[data_indice]--;
      break;
    case '>':
      data_indice++;
      break;
    case '<':
      data_indice--;
      break;
    case '.':
      putchar(data[data_indice]);
      break;
    case ',':
      data[data_indice] = getchar();
      break;
    case '[': {
      if (data[data_indice] != 0)
        *(p_stack++) = code_ptr;
      else {
        level = 1;
        do {
          code_ptr++;
          if (*code_ptr == '[')
            level++;
          else if (*code_ptr == ']')
            level--;
        } while (level);
      }
    } break;
    case ']':
      code_ptr = *(--p_stack) - 1;
      break;
    default:
      break;
    }
    code_ptr++;
  }
}
```