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
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x2c(%rbp)
  jmp    L89
L86:
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x2c(%rbp)
L89:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L86
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x1,-0x2c(%rbp)
  jmp    L82
L78:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,-0x28(%rbp)
  jge    L98
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %edx,-0x20(%rbp,%rax,4)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x20(%rbp)
L98:
  movl   $0x1,-0x2c(%rbp)
  jmp    L54
L27:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,-0x24(%rbp)
  jle    L44
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %edx,-0x20(%rbp,%rax,4)
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x24(%rbp),%edx
  mov    %edx,-0x20(%rbp,%rax,4)
L44:
  addl   $0x1,-0x2c(%rbp)
L54:
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L27
  addl   $0x1,-0x2c(%rbp)
L82:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L78
  movl   $0x0,-0x2c(%rbp)
  jmp    L45
L88:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x2c(%rbp)
L45:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L88
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L70
  call   <__stack_chk_fail@plt>
L70:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D37
datatype: string
name: extract_symbolic_data
arguments: data_tag: D35
datatype: string
name: extract_symbolic_data
arguments: data_tag: D46
datatype: string

tool: D37: Nhap so phan tu cho chuoi : 

tool: D35: %d

tool: D46: Nhap vao phan tu thu %d cho chuoi : 

assistant: ```c
int main() {
  int arr[5], i, N;
  printf("Nhap so phan tu cho chuoi : ");
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    printf("Nhap vao phan tu thu %d cho chuoi : ", i + 1);
    scanf("%d", &arr[i]);
  }
  int imax = arr[0];
  int imin = arr[N - 1];
  for (i = 1; i <= N - 1; i++) {
    if (imax < arr[i]) {
      imax = arr[i];
      arr[i] = arr[0];
      arr[0] = imax;
    }
    for (i = 1; i < N - 1; i++)
      if (imin > arr[i]) {
        imin = arr[i];
        arr[i] = arr[N - 1];
        arr[N - 1] = imin;
      }
  }
  for (i = 0; i < N; i++) {
    printf("%d", arr[i]);
  }
}
```