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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D79(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  mov    -0x20(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  mov    -0x20(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  mov    -0x20(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D28
type_infered: string
name: parse_data
arguments: data_label: D48
type_infered: string
name: parse_data
arguments: data_label: D77
type_infered: string
name: parse_data
arguments: data_label: D79
type_infered: string
name: parse_data
arguments: data_label: D58
type_infered: string
name: parse_data
arguments: data_label: D0
type_infered: string
name: parse_data
arguments: data_label: D33
type_infered: string
name: parse_data
arguments: data_label: D89
type_infered: string

tool: D28: what is the first integer

tool: D48: %d

tool: D77: what is the second integer

tool: D79: the sum for the two integers is:%d 


tool: D58: the difference for the two integers is:%d


tool: D0: the product for the two integers is: %d


tool: D33: the quotient for the two integers is:%d


tool: D89: the modulus for the two integers is:%d


assistant: ```c
int main() {
  int first;
  printf("what is the first integer\n");
  scanf("%d", &first);
  int second;
  printf("what is the second integer\n");
  scanf("%d", &second);
  int sum = (int)first + second;
  printf("the sum for the two integers is:%d \n", sum);
  int difference = (int)first - second;
  printf("the difference for the two integers is:%d\n", difference);
  int product = (int)first * second;
  printf("the product for the two integers is: %d\n", product);
  int quotient = (int)first / second;
  printf("the quotient for the two integers is:%d\n", quotient);
  int modulus = (int)first % second;
  printf("the modulus for the two integers is:%d\n", modulus);
  return 0;
}
```