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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L7
L58:
  mov    -0x24(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x7(%rax),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  cmpl   $0x3,-0x20(%rbp)
  je     L14
  cmpl   $0x3,-0x20(%rbp)
  jg     L24
  cmpl   $0x2,-0x20(%rbp)
  je     L39
  cmpl   $0x2,-0x20(%rbp)
  jg     L24
  cmpl   $0x0,-0x20(%rbp)
  je     L23
  cmpl   $0x1,-0x20(%rbp)
  je     L0
  jmp    L24
L23:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L24
L0:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L24
L39:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L24
L14:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L24:
  addl   $0x1,-0x20(%rbp)
L7:
  cmpl   $0x3,-0x20(%rbp)
  jle    L58
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L112
  call   <__stack_chk_fail@plt>
L112:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D27
data_type: string
name: retrieve_data
arguments: data_tag: D8
data_type: string
name: retrieve_data
arguments: data_tag: D21
data_type: string

tool: D27: Enter a four-digit integer number:

tool: D8: %d

tool: D21: The encrypted number is :%d%d%d%d

assistant: ```c
int main() {
  int counter, number, digit = 0;
  int first_digit, second_digit, third_digit, fourth_digit;
  printf("Enter a four-digit integer number:");
  scanf("%d", &number);
  for (counter = 0; counter < 4; counter++) {
    digit = number % 10;
    number = number / 10;
    digit = (digit + 7) % 10;
    switch (counter) {
    case 0:
      third_digit = digit;
      break;
    case 1:
      fourth_digit = digit;
      break;
    case 2:
      first_digit = digit;
      break;
    case 3:
      second_digit = digit;
    }
  }
  printf("The encrypted number is :%d%d%d%d", fourth_digit, third_digit, second_digit, first_digit);
  return 0;
}
```