user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x19(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl -0x19(%rbp),%eax
  movsbl %al,%eax
  cmp    $0x2f,%eax
  je     L23
  cmp    $0x2f,%eax
  jg     L76
  cmp    $0x2d,%eax
  je     L27
  cmp    $0x2d,%eax
  jg     L76
  cmp    $0x2a,%eax
  je     L85
  cmp    $0x2b,%eax
  jne    L76
  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    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L75
L27:
  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    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L75
L85:
  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    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L75
L23:
  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    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L75
L76:
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L75:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L89
  call   <__stack_chk_fail@plt>
L89:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D49
data_type: string
name: unpack_data
arguments: label: D35
data_type: string
name: unpack_data
arguments: label: D80
data_type: string
name: unpack_data
arguments: label: D42
data_type: string
name: unpack_data
arguments: label: D47
data_type: string
name: unpack_data
arguments: label: D37
data_type: string
name: unpack_data
arguments: label: D18
data_type: string
name: unpack_data
arguments: label: D56
data_type: string
name: unpack_data
arguments: label: D0
data_type: string

tool: D49: Enter an operator (+, -, *,/): 

tool: D35: %c

tool: D80: Enter two numbers: 

tool: D42: %lf %lf

tool: D47: The result of the operation %.1lf + %.1lf = %.1lf

tool: D37: The result of the operation %.1lf - %.1lf = %.1lf

tool: D18: The result of the operation %.1lf * %.1lf = %.1lf

tool: D56: The result of the operation %.1lf / %.1lf = %.1lf

tool: D0: Error! The entered operator is invalid

assistant: ```c
int main() {
  char op;
  double num1, num2;
  printf("Enter an operator (+, -, *,/): ");
  scanf("%c", &op);
  printf("Enter two numbers: ");
  scanf("%lf %lf", &num1, &num2);
  switch (op) {
  case '+':
    printf("The result of the operation %.1lf + %.1lf = %.1lf", num1, num2, num1 + num2);
    break;
  case '-':
    printf("The result of the operation %.1lf - %.1lf = %.1lf", num1, num2, num1 - num2);
    break;
  case '*':
    printf("The result of the operation %.1lf * %.1lf = %.1lf", num1, num2, num1 * num2);
    break;
  case '/':
    printf("The result of the operation %.1lf / %.1lf = %.1lf", num1, num2, num1 / num2);
    break;
  default:
    printf("Error! The entered operator is invalid");
  }
  return 0;
}
```