user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D117(%rip),%r15
  push   %r14
  push   %r13
  lea    D87(%rip),%r13
  push   %r12
  lea    D49(%rip),%r12
  push   %rbp
  lea    D44(%rip),%rbp
  push   %rbx
  lea    D26(%rip),%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  movl   $0x1,0x2c(%rsp)
  lea    0x2b(%rsp),%r14
  mov    %rax,0x10(%rsp)
  lea    0x34(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x2c(%rsp),%rax
  mov    %rax,0x8(%rsp)
L14:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D67(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movzbl 0x2b(%rsp),%eax
  sub    $0x2a,%eax
  cmp    $0xa,%al
  ja     L30
  movzbl %al,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  nopl   0x0(%rax,%rax,1)
  mov    0x30(%rsp),%esi
  mov    0x34(%rsp),%r8d
  lea    D81(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r8d
  mov    %r8d,%edx
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
L59:
  xor    %esi,%esi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
  mov    0x30(%rsp),%esi
  mov    0x34(%rsp),%edx
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  mov    %esi,%ecx
  sub    %edx,%ecx
  call   <printf@plt>
  jmp    L59
  nopl   0x0(%rax)
  mov    0x34(%rsp),%edx
  mov    0x30(%rsp),%esi
  lea    D107(%rip),%rdi
  xor    %eax,%eax
  lea    (%rdx,%rsi,1),%ecx
  call   <printf@plt>
  jmp    L59
  nopl   0x0(%rax,%rax,1)
  mov    0x30(%rsp),%esi
  mov    0x34(%rsp),%edx
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  mov    %esi,%ecx
  imul   %edx,%ecx
  call   <printf@plt>
  jmp    L59
L30:
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L59
```

tool_calls: name: read_data
arguments: data_tag: D117
type_infered: string
name: read_data
arguments: data_tag: D87
type_infered: string
name: read_data
arguments: data_tag: D49
type_infered: string
name: read_data
arguments: data_tag: D26
type_infered: string
name: read_data
arguments: data_tag: D67
type_infered: string
name: read_data
arguments: data_tag: D81
type_infered: string
name: read_data
arguments: data_tag: D24
type_infered: string
name: read_data
arguments: data_tag: D63
type_infered: string
name: read_data
arguments: data_tag: D107
type_infered: string
name: read_data
arguments: data_tag: D38
type_infered: string
name: read_data
arguments: data_tag: D68
type_infered: string

tool: D117: op

tool: D87: %c

tool: D49: num1

tool: D26: %d

tool: D67: num2

tool: D81:  %d / %d = %d

tool: D24: %d


tool: D63:  %d - %d = %d

tool: D107:  %d + %d = %d

tool: D38:  %d * %d = %d

tool: D68: error

assistant: ```c
main() {
  int l = 1;
  while ("l==1") {
    char op;
    int num1, num2, num3;
    printf("op");
    scanf("%c", &op);
    printf("num1");
    scanf("%d", &num1);
    printf("num2");
    scanf("%d", &num2);
    switch (op) {
    case '1':
    case '+':
      printf(" %d + %d = %d", num1, num2, num2 + num1);
      break;
    case '2':
    case '-':
      printf(" %d - %d = %d", num1, num2, num1 - num2);
      break;
    case '3':
    case '*':
      printf(" %d * %d = %d", num1, num2, num1 * num2);
      break;
    case '4':
    case '/':
      printf(" %d / %d = %d", num1, num2, num1 / num2);
      break;
    default:
      printf("error");
      break;
    }
    printf("%d\n", num3);
    scanf("%d", &l);
  }
}
```