user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rcx
  lea    -0x19(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movzbl -0x19(%rbp),%eax
  movsbl %al,%eax
  sub    $0x25,%eax
  cmp    $0xa,%eax
  ja     L31
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D103(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D103(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L26
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L26
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L26
  movsd  -0x10(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L5
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L5
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L26
L5:
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L26
  movsd  -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%rax
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  mov    %rax,%rsi
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L26
L31:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L26:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D62
type_guessed: string
name: read_data
arguments: label: D21
type_guessed: string
name: read_data
arguments: label: D9
type_guessed: string
name: read_data
arguments: label: D57
type_guessed: string
name: read_data
arguments: label: D63
type_guessed: string
name: read_data
arguments: label: D24
type_guessed: string

tool: D62: Enter the calculation:

tool: D21: %lf%c%lf

tool: D9: =%lf


tool: D57: Division by zero error!

tool: D63: =%l


tool: D24: Invalid Operator

assistant: ```c
void main() {
  double num1, num2;
  char op;
  printf("Enter the calculation:\n");
  scanf("%lf%c%lf", &num1, &op, &num2);
  switch (op) {
  case '+':
    printf("=%lf\n", num1 + num2);
    break;
  case '-':
    printf("=%lf\n", num1 - num2);
    break;
  case '*':
    printf("=%lf\n", num1 * num2);
    break;
  case '/':
    if (num2 == 0) {
      printf("Division by zero error!\n");
    } else {
      printf("=%lf\n", num1 / num2);
    }
    break;
  case '%':
    printf("=%l\n", (long)num1 % (long)num2);
    break;
  default:
    printf("Invalid Operator\n");
  }
}
```