user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  lea    D56(%rip),%rdi
  push   %r14
  push   %r13
  lea    D45(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0xfc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%r15
  lea    0x10(%rsp),%rbx
  call   <puts@plt>
  lea    D57(%rip),%rdi
  call   <puts@plt>
  lea    D67(%rip),%rdi
  call   <puts@plt>
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x1,0xc(%rsp)
  je     L8
L57:
  cmpl   $0x2,0xc(%rsp)
  je     L91
L33:
  lea    -0x2(%rbp),%eax
  lea    0x14(%rsp,%rax,4),%rsi
L105:
  cmp    $0x1,%ebp
  jle    L2
  mov    %rbx,%rax
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L46:
  mov    (%rax),%ecx
  mov    0x4(%rax),%edx
  cmp    %edx,%ecx
  jle    L76
  mov    %ecx,0x4(%rax)
  mov    $0x1,%edi
  mov    %edx,(%rax)
L76:
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L46
  test   %edi,%edi
  jne    L105
L2:
  movl   $0x0,0xc(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rbx
  nopl   0x0(%rax)
L24:
  cltq
  mov    %rbx,%rdi
  mov    0x10(%rsp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  add    $0x1,%eax
  mov    %eax,0xc(%rsp)
  cmp    %ebp,%eax
  jne    L24
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0xfc8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    %rbx,%r12
  lea    D35(%rip),%r14
  cs nopw 0x0(%rax,%rax,1)
L27:
  add    $0x1,%ebp
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <scanf@plt>
  cmpl   $0x2a,-0x4(%r12)
  jne    L27
  jmp    L57
L91:
  lea    D52(%rip),%rdi
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  xor    %r12d,%r12d
  call   <printf@plt>
  mov    %r15,%rsi
  mov    %r13,%rdi
  mov    %rbx,%r15
  call   <scanf@plt>
  lea    D35(%rip),%r14
  nopl   0x0(%rax)
L119:
  add    $0x1,%r12d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %r12d,%esi
  call   <printf@plt>
  mov    %r15,%rsi
  mov    %r13,%rdi
  add    $0x4,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%ebp
  cmp    %ebp,%r12d
  jne    L119
  jmp    L33
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D56
datatype: string
name: unpack_data
arguments: tag: D45
datatype: string
name: unpack_data
arguments: tag: D57
datatype: string
name: unpack_data
arguments: tag: D67
datatype: string
name: unpack_data
arguments: tag: D28
datatype: string
name: unpack_data
arguments: tag: D35
datatype: string
name: unpack_data
arguments: tag: D52
datatype: string

tool: D56: Choose when to stop(Insert 1 or 2):

tool: D45: %d

tool: D57: 1.When enter number 42!

tool: D67: 2.When you want!

tool: D28: %d 

tool: D35: Insert %d number:

tool: D52: Insert number of numbers:

assistant: ```c
int main() {
  int i = 0, n = 0, p = 0;
  int a[1000];
  printf("Choose when to stop(Insert 1 or 2):\n");
  printf("1.When enter number 42!\n");
  printf("2.When you want!\n");
  scanf("%d", &n);
  if (n == 1) {
    do {
      printf("Insert %d number:", i + 1);
      scanf("%d", &a[i]);
      i++;
    } while (a[i - 1] != 42);
  }
  if (n == 2) {
    i = 0;
    n = 0;
    printf("Insert number of numbers:");
    scanf("%d", &n);
    do {
      printf("Insert %d number:", i + 1);
      scanf("%d", &a[i]);
      i++;
    } while (i != n);
  }
  n = 0;
  while (1) {
    p = 0;
    for (n = 0; n < i - 1; n++) {
      if (a[n] > a[n + 1]) {
        a[n] = a[n] + a[n + 1];
        a[n + 1] = a[n] - a[n + 1];
        a[n] = a[n] - a[n + 1];
        p = 1;
      }
    }
    if (p == 0) {
      break;
    }
  }
  n = 0;
  do {
    printf("%d ", a[n]);
    n++;
  } while (n != i);
  printf("\n");
}
```