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    $0x70,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L89
L195:
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <ungets@plt>
  addq   $0x8,-0x60(%rbp)
  mov    -0x60(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <ungets@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <getop@plt>
  cmp    $0x6d,%eax
  je     L113
  cmp    $0x6d,%eax
  jg     L58
  cmp    $0x64,%eax
  je     L147
  cmp    $0x64,%eax
  jg     L58
  cmp    $0x2d,%eax
  je     L78
  cmp    $0x2d,%eax
  jg     L58
  test   %eax,%eax
  je     L83
  cmp    $0x2b,%eax
  je     L183
  jmp    L58
L83:
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <push@plt>
  jmp    L89
L183:
  mov    $0x0,%eax
  call   <pop@plt>
  movsd  %xmm0,-0x68(%rbp)
  mov    $0x0,%eax
  call   <pop@plt>
  movapd %xmm0,%xmm1
  addsd  -0x68(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <push@plt>
  jmp    L89
L78:
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x48(%rbp)
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm2
  subsd  -0x48(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  call   <push@plt>
  jmp    L89
L113:
  mov    $0x0,%eax
  call   <pop@plt>
  movsd  %xmm0,-0x68(%rbp)
  mov    $0x0,%eax
  call   <pop@plt>
  movapd %xmm0,%xmm3
  mulsd  -0x68(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  call   <push@plt>
  jmp    L89
L147:
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x48(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x48(%rbp),%xmm0
  jp     L10
  pxor   %xmm0,%xmm0
  ucomisd -0x48(%rbp),%xmm0
  je     L188
L10:
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm4
  divsd  -0x48(%rbp),%xmm4
  movq   %xmm4,%rax
  movq   %rax,%xmm0
  call   <push@plt>
  jmp    L89
L188:
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L89
L58:
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x54(%rbp)
  nop
L89:
  subl   $0x1,-0x54(%rbp)
  cmpl   $0x0,-0x54(%rbp)
  jg     L195
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
data_type: string
name: retrieve_data
arguments: data_label: D30
data_type: string
name: retrieve_data
arguments: data_label: D21
data_type: string
name: retrieve_data
arguments: data_label: D11
data_type: string

tool: D9:  

tool: D30: math error: divide by 0

tool: D21: error: unindentified character %s


tool: D11: result is %f


assistant: ```c
int main(int argc, char *argv[]) {
  double op2;
  char s[50];
  while (--argc > 0) {
    ungets(" ");
    ungets(*++argv);
    switch (getop(s)) {
    case 0:
      push(atof(s));
      break;
    case '+':
      push(pop() + pop());
      break;
    case '-':
      op2 = pop();
      push(pop() - op2);
      break;
    case 'm':
      push(pop() * pop());
      break;
    case 'd':
      op2 = pop();
      if (op2 != 0.0)
        push(pop() / op2);
      else
        printf("math error: divide by 0\n");
      break;
    default:
      printf("error: unindentified character %s\n", s);
      argc = 1;
      break;
    }
  }
  printf("result is %f\n", pop());
  return 0;
}
```