user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x400,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3fc(%rbp)
  movl   $0x0,-0x3f8(%rbp)
  jmp    L1
L39:
  movl   $0x0,-0x400(%rbp)
  jmp    L152
L113:
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x22,%al
  jne    L95
  movl   $0x1,-0x3f8(%rbp)
L95:
  cmpl   $0x1,-0x3f8(%rbp)
  jne    L20
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x22,%al
  jne    L20
  movl   $0x0,-0x3f8(%rbp)
L20:
  cmpl   $0x0,-0x3f8(%rbp)
  jne    L3
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2f,%al
  jne    L9
  mov    -0x400(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2f,%al
  jne    L9
  addl   $0x2,-0x400(%rbp)
  jmp    L13
L194:
  addl   $0x1,-0x400(%rbp)
L13:
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L194
L9:
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2f,%al
  jne    L84
  mov    -0x400(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2a,%al
  jne    L84
  addl   $0x2,-0x400(%rbp)
  movl   $0x1,-0x3fc(%rbp)
L84:
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2a,%al
  jne    L135
  mov    -0x400(%rbp),%eax
  add    $0x1,%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  cmp    $0x2f,%al
  jne    L135
  addl   $0x2,-0x400(%rbp)
  movl   $0x0,-0x3fc(%rbp)
L135:
  cmpl   $0x1,-0x3fc(%rbp)
  jne    L176
  addl   $0x1,-0x400(%rbp)
  jmp    L152
L176:
  cmpl   $0x0,-0x3fc(%rbp)
  jne    L152
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x400(%rbp)
  jmp    L152
L3:
  cmpl   $0x1,-0x3f8(%rbp)
  jne    L152
  mov    -0x400(%rbp),%eax
  cltq
  movzbl -0x3f0(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x400(%rbp)
L152:
  mov    -0x400(%rbp),%eax
  cmp    -0x3f4(%rbp),%eax
  jl     L113
L1:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getlineuser@plt>
  mov    %eax,-0x3f4(%rbp)
  cmpl   $0x0,-0x3f4(%rbp)
  jg     L39
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L109
  call   <__stack_chk_fail@plt>
L109:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D12
type_guessed: string

tool: D12: %c

assistant: ```c
int main() {
  int len, t, flag_comment, flag_quote;
  char line[1000];
  flag_comment = 0;
  flag_quote = 0;
  while ((len = getlineuser(line, 1000)) > 0) {
    t = 0;
    while (t < len) {
      if (line[t] == '"') {
        flag_quote = 1;
      }
      if (flag_quote == 1 && line[t] == '"') {
        flag_quote = 0;
      }
      if (flag_quote == 0) {
        if (line[t] == '/' && line[t + 1] == '/') {
          t = t + 2;
          while (line[t] != '\n')
            t++;
        }
        if (line[t] == '/' && line[t + 1] == '*') {
          t = t + 2;
          flag_comment = 1;
        }
        if (line[t] == '*' && line[t + 1] == '/') {
          t = t + 2;
          flag_comment = 0;
        }
        if (flag_comment == 1)
          t++;
        else if (flag_comment == 0) {
          printf("%c", line[t]);
          t++;
        }
      } else if (flag_quote == 1) {
        printf("%c", line[t]);
        t++;
      }
    }
  }
  return 0;
}
```