user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D70(%rip),%rdi
  push   %r14
  push   %r13
  lea    D45(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfc8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x1c(%rsp)
  call   <puts@plt>
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D73(%rip),%rdi
  call   <puts@plt>
  lea    0x1c(%rsp),%rax
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,0x8(%rsp)
  call   <scanf@plt>
  cmpl   $0x1,0x1c(%rsp)
  je     L90
  or     $0xffffffff,%r12d
  xor    %ebp,%ebp
  lea    0x20(%rsp),%rbx
L15:
  cmpl   $0x2,0x1c(%rsp)
  je     L73
L29:
  lea    -0x1(%r12),%eax
  lea    0x24(%rsp,%rax,4),%rsi
  nopl   0x0(%rax,%rax,1)
L39:
  test   %r12d,%r12d
  jle    L13
  mov    %rbx,%rax
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L21:
  mov    (%rax),%ecx
  mov    0x4(%rax),%edx
  cmp    %edx,%ecx
  jle    L16
  mov    %ecx,0x4(%rax)
  mov    $0x1,%edi
  mov    %edx,(%rax)
L16:
  add    $0x4,%rax
  cmp    %rax,%rsi
  jne    L21
  test   %edi,%edi
  jne    L39
L13:
  movl   $0x0,0x1c(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rbx
  nopl   0x0(%rax)
L3:
  cltq
  mov    %rbx,%rdi
  mov    0x20(%rsp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x1c(%rsp),%eax
  add    $0x1,%eax
  mov    %eax,0x1c(%rsp)
  cmp    %ebp,%eax
  jne    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L78
  add    $0xfd8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L90:
  lea    0x20(%rsp),%rbx
  xor    %ebp,%ebp
  lea    D33(%rip),%r14
  mov    %rbx,%r15
  nopl   (%rax)
L40:
  mov    %ebp,%r12d
  add    $0x1,%ebp
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %r15,%rsi
  mov    %r13,%rdi
  add    $0x4,%r15
  call   <scanf@plt>
  cmpl   $0x2a,-0x4(%r15)
  jne    L40
  jmp    L15
L73:
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  mov    %rbx,%r15
  xor    %r12d,%r12d
  movl   $0x0,0x1c(%rsp)
  lea    D33(%rip),%r14
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  nopl   (%rax)
L98:
  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    0x1c(%rsp),%ebp
  cmp    %ebp,%r12d
  jne    L98
  lea    -0x1(%rbp),%r12d
  jmp    L29
L78:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D70
type_guessed: string
name: read_data
arguments: tag: D45
type_guessed: string
name: read_data
arguments: tag: D22
type_guessed: string
name: read_data
arguments: tag: D73
type_guessed: string
name: read_data
arguments: tag: D10
type_guessed: string
name: read_data
arguments: tag: D33
type_guessed: string
name: read_data
arguments: tag: D31
type_guessed: string

tool: D70: Choose when to stop(Insert 1 or 2):

tool: D45: %d

tool: D22: 1.When enter number 42!

tool: D73: 2.When you want!

tool: D10: %d 

tool: D33: Insert %d number:

tool: D31: 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");
}
```