user: 请从给定的汇编指令推导出C源代码：

```asm
<libbf_naive_interpreter_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x60000(%rsp),%r11
L86:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L86
  sub    $0x40,%rsp
  mov    %rdi,-0x60038(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x60038(%rbp),%rax
  mov    %rax,-0x60028(%rbp)
  lea    -0x60010(%rbp),%rax
  mov    $0x20000,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movq   $0x0,-0x60020(%rbp)
  lea    -0x40010(%rbp),%rax
  mov    %rax,-0x60018(%rbp)
L134:
  mov    -0x60028(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  test   %eax,%eax
  je     L185
  test   %eax,%eax
  js     L32
  cmp    $0x5d,%eax
  jg     L32
  cmp    $0x2b,%eax
  jl     L32
  sub    $0x2b,%eax
  cmp    $0x32,%eax
  ja     L32
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D11(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D11(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x60020(%rbp),%rax
  mov    -0x60010(%rbp,%rax,4),%eax
  lea    0x1(%rax),%edx
  mov    -0x60020(%rbp),%rax
  mov    %edx,-0x60010(%rbp,%rax,4)
  jmp    L158
  mov    -0x60020(%rbp),%rax
  mov    -0x60010(%rbp,%rax,4),%eax
  lea    -0x1(%rax),%edx
  mov    -0x60020(%rbp),%rax
  mov    %edx,-0x60010(%rbp,%rax,4)
  jmp    L158
  addq   $0x1,-0x60020(%rbp)
  jmp    L158
  subq   $0x1,-0x60020(%rbp)
  jmp    L158
  mov    -0x60020(%rbp),%rax
  mov    -0x60010(%rbp,%rax,4),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L158
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    -0x60020(%rbp),%rdx
  mov    %eax,-0x60010(%rbp,%rdx,4)
  jmp    L158
  mov    -0x60020(%rbp),%rax
  mov    -0x60010(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L192
  mov    -0x60018(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x60018(%rbp)
  mov    -0x60028(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L158
L192:
  movl   $0x1,-0x6002c(%rbp)
L19:
  addq   $0x1,-0x60028(%rbp)
  mov    -0x60028(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5b,%al
  jne    L203
  addl   $0x1,-0x6002c(%rbp)
  jmp    L104
L203:
  mov    -0x60028(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5d,%al
  jne    L104
  subl   $0x1,-0x6002c(%rbp)
L104:
  cmpl   $0x0,-0x6002c(%rbp)
  jne    L19
  jmp    L158
  subq   $0x8,-0x60018(%rbp)
  mov    -0x60018(%rbp),%rax
  mov    (%rax),%rax
  sub    $0x1,%rax
  mov    %rax,-0x60028(%rbp)
  jmp    L158
L32:
  nop
L158:
  addq   $0x1,-0x60028(%rbp)
  jmp    L134
L185:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

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++;
  }
}
```