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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jge    L30
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L30
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L30:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L62
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jge    L62
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L62:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L1
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L1
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L1:
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%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     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D74
type_guessed: string
name: parse_data
arguments: data_label: D65
type_guessed: string
name: parse_data
arguments: data_label: D16
type_guessed: string
name: parse_data
arguments: data_label: D49
type_guessed: string
name: parse_data
arguments: data_label: D48
type_guessed: string
name: parse_data
arguments: data_label: D43
type_guessed: string
name: parse_data
arguments: data_label: D89
type_guessed: string

tool: D74: Riordinatore di tre numeri

tool: D65: Inserisci il primo numero: 

tool: D16: %d

tool: D49: Inserisci il secondo numero: 

tool: D48: Inserisci il terzo numero: 

tool: D43: Numeri riordinati in ordine crescente: %d %d %d

tool: D89: Numeri riordinati in ordine decrescente: %d %d %d

assistant: ```c
int main() {
  int x, y, z;
  int temp;
  printf("Riordinatore di tre numeri\n");
  printf("Inserisci il primo numero: ");
  scanf("%d", &x);
  printf("Inserisci il secondo numero: ");
  scanf("%d", &y);
  printf("Inserisci il terzo numero: ");
  scanf("%d", &z);
  if (x < y && y > z) {
    temp = z;
    z = y;
    y = temp;
  }
  if (x > y && y < z) {
    temp = x;
    x = y;
    y = temp;
  }
  if (x > y && y > z) {
    temp = x;
    x = z;
    z = temp;
  }
  printf("Numeri riordinati in ordine crescente: %d %d %d", x, y, z);
  printf("Numeri riordinati in ordine decrescente: %d %d %d", z, y, x);
  return 0;
}
```