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    D110(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x19(%rbp),%rax
  mov    %rax,%rsi
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D90(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl -0x19(%rbp),%eax
  movsbl %al,%eax
  sub    $0x25,%eax
  cmp    $0xa,%eax
  ja     L33
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D27(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D27(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  addsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D107(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L31
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  movapd %xmm0,%xmm2
  subsd  %xmm1,%xmm2
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D100(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L31
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L31
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L31
  movsd  -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%ecx
  cltd
  idiv   %ecx
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L31
L33:
  lea    D112(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L31:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D110
data_type: string
name: disassemble_data
arguments: data_label: D66
data_type: string
name: disassemble_data
arguments: data_label: D68
data_type: string
name: disassemble_data
arguments: data_label: D90
data_type: string
name: disassemble_data
arguments: data_label: D107
data_type: string
name: disassemble_data
arguments: data_label: D100
data_type: string
name: disassemble_data
arguments: data_label: D81
data_type: string
name: disassemble_data
arguments: data_label: D2
data_type: string
name: disassemble_data
arguments: data_label: D86
data_type: string
name: disassemble_data
arguments: data_label: D112
data_type: string

tool: D110: Which operation you want to perform (+,-,*,/,%%)?


tool: D66: %c

tool: D68: Enter two operands:

tool: D90: %lf%lf

tool: D107: addition of %0.2lf & %0.2lf is: %0.4lf


tool: D100: subtraction of %0.2lf & %0.2lf is: %0.4lf


tool: D81: multiplication of %0.2lf & %0.2lf is: %0.4lf


tool: D2: division of %0.2lf & %0.2lf is: %0.4lf


tool: D86: mod remainder of %0.2lf & %0.2lf is: %d


tool: D112: Please enter valid operator

assistant: ```c
void main(void) {
  double a, b;
  char operator;
  printf("Which operation you want to perform (+,-,*,/,%%)?\n");
  scanf("%c", &operator);
  printf("Enter two operands:\n");
  scanf("%lf%lf", &a, &b);
  switch (operator) {
  case '+':
    printf("addition of %0.2lf & %0.2lf is: %0.4lf\n", a, b, a + b);
    break;
  case '-':
    printf("subtraction of %0.2lf & %0.2lf is: %0.4lf\n", a, b, a - b);
    break;
  case '*':
    printf("multiplication of %0.2lf & %0.2lf is: %0.4lf\n", a, b, a * b);
    break;
  case '/':
    printf("division of %0.2lf & %0.2lf is: %0.4lf\n", a, b, a / b);
    break;
  case '%':
    printf("mod remainder of %0.2lf & %0.2lf is: %d\n", a, b, (int)a % (int)b);
    break;
  default:
    printf("Please enter valid operator\n");
  }
}
```