user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D39(%rip),%r12
  lea    D51(%rip),%rbp
  lea    D89(%rip),%rbx
  lea    D66(%rip),%r13
  jmp    L44
L4:
  lea    D76(%rip),%rdi
  call   <puts@plt>
L39:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  pxor   %xmm0,%xmm0
  ucomiss 0x24(%rsp),%xmm0
  jp     L13
  je     L49
L13:
  movss  0x20(%rsp),%xmm0
  divss  0x24(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  jmp    L22
L49:
  lea    D103(%rip),%rdi
  call   <puts@plt>
  jmp    L39
  movss  0x20(%rsp),%xmm0
  addss  0x24(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
L22:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  lea    D25(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  cmpb   $0x31,0x1f(%rsp)
  jne    L39
L44:
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  lea    0x1f(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpb   $0x32,0x1f(%rsp)
  je     L4
  mov    %r13,%rdi
  call   <puts@plt>
  lea    D55(%rip),%rdi
  call   <puts@plt>
  lea    0x24(%rsp),%rcx
  lea    0x1e(%rsp),%rdx
  lea    0x20(%rsp),%rsi
  lea    D67(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl 0x1e(%rsp),%eax
  cmp    $0x2f,%al
  je     L18
  sub    $0x25,%eax
  cmp    $0xa,%al
  ja     L70
  movzbl %al,%eax
  lea    D81(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  movss  0x20(%rsp),%xmm0
  subss  0x24(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  jmp    L22
  movss  0x20(%rsp),%xmm0
  mulss  0x24(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  jmp    L22
  cvttss2si 0x20(%rsp),%eax
  cvttss2si 0x24(%rsp),%ecx
  cltd
  idiv   %ecx
  pxor   %xmm1,%xmm1
  cvtsi2ss %edx,%xmm1
  movss  %xmm1,0xc(%rsp)
  jmp    L22
L70:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L22
L92:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D39
type_guessed: string
name: unpack_data
arguments: tag: D51
type_guessed: string
name: unpack_data
arguments: tag: D89
type_guessed: string
name: unpack_data
arguments: tag: D66
type_guessed: string
name: unpack_data
arguments: tag: D76
type_guessed: string
name: unpack_data
arguments: tag: D103
type_guessed: string
name: unpack_data
arguments: tag: D25
type_guessed: string
name: unpack_data
arguments: tag: D55
type_guessed: string
name: unpack_data
arguments: tag: D67
type_guessed: string
name: unpack_data
arguments: tag: D4
type_guessed: string

tool: D39: (1) Calcular

tool: D51: (2) Sair

tool: D89:  %c

tool: D66: Escreva o numero na seguinte forma 	 N1 * N2

tool: D76: SAINDO

tool: D103: Operacao nao permitida, SAINDO

tool: D25: %f


tool: D55: Digite aqui o numero

tool: D67: %f %c %f

tool: D4: SINAL ERRADO

assistant: ```c
int main() {
  float result;
  char sinal, continuar;
  float n1, n2;
  do {
    printf("(1) Calcular\n");
    printf("(2) Sair\n");
    scanf(" %c", &continuar);
    if (continuar == '2') {
      printf("SAINDO\n");
      break;
    }
    printf("Escreva o numero na seguinte forma \t N1 * N2\n");
    printf("Digite aqui o numero\n");
    scanf("%f %c %f", &n1, &sinal, &n2);
    if ((sinal == '/') && (n2 == 0)) {
      printf("Operacao nao permitida, SAINDO\n");
      return 0;
    } else {
      switch (sinal) {
      case '+':
        result = n1 + n2;
        break;
      case '-':
        result = n1 - n2;
        break;
      case '*':
        result = n1 * n2;
        break;
      case '/':
        result = n1 / n2;
        break;
      case '%':
        result = (int)n1 % (int)n2;
        break;
      default:
        printf("SINAL ERRADO\n");
      }
      printf("%f\n", result);
    }
  } while (continuar == '1');
  return 0;
}
```