user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,0xc(%rsp)
  movl   $0x8,0x10(%rsp)
  mov    $0x8,%edx
  mov    $0x5,%esi
  lea    D118(%rip),%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  mov    0xc(%rsp),%edi
  call   <exchange@plt>
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  lea    0xc(%rsp),%rdi
  call   <exchange_ptr@plt>
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D65(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D70(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D91(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%edi
  call   <allocate_ints@plt>
  test   %rax,%rax
  je     L7
  mov    %rax,%rbp
  lea    D16(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x14(%rsp)
  je     L41
  mov    $0x0,%ebx
  lea    D79(%rip),%r14
  lea    D93(%rip),%r13
L1:
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%r12d
  add    $0x1,%ebx
  mov    %r12d,%eax
  lea    0x0(%rbp,%rax,4),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmp    %ebx,0x14(%rsp)
  ja     L1
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebx,%ebx
  je     L65
  mov    %r12d,%r12d
  lea    0x0(%rbp,%r12,4),%rbx
  lea    D35(%rip),%r12
L5:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rax
  sub    $0x4,%rbx
  cmp    %rbp,%rax
  jne    L5
L65:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
L11:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L66
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  lea    D49(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%eax
  jmp    L11
L41:
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L65
L66:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D118
type_infered: string
name: unpack_data
arguments: data_label: D6
type_infered: string
name: unpack_data
arguments: data_label: D65
type_infered: string
name: unpack_data
arguments: data_label: D70
type_infered: string
name: unpack_data
arguments: data_label: D91
type_infered: string
name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D79
type_infered: string
name: unpack_data
arguments: data_label: D93
type_infered: string
name: unpack_data
arguments: data_label: D20
type_infered: string
name: unpack_data
arguments: data_label: D35
type_infered: string
name: unpack_data
arguments: data_label: D49
type_infered: string

tool: D118: first:
x = %d, y = %d



tool: D6: exchange:
x = %d, y = %d



tool: D65: exchange_ptr:
x = %d, y = %d



tool: D70: How many elements would you like to store? 

tool: D91: %u

tool: D16: Enter data:

tool: D79: %u = 

tool: D93: %d

tool: D20: The stored data is as follows:

tool: D35:  %d

tool: D49: Cannot allocate array of that size.

assistant: ```c
int main() {
  int x = 5, y = 8;
  printf("first:\nx = %d, y = %d\n\n", x, y);
  exchange(x, y);
  printf("exchange:\nx = %d, y = %d\n\n", x, y);
  exchange_ptr(&x, &y);
  printf("exchange_ptr:\nx = %d, y = %d\n\n", x, y);
  unsigned size;
  printf("How many elements would you like to store? ");
  scanf("%u", &size);
  int *nums = allocate_ints(size);
  if (!nums) {
    perror("Cannot allocate array of that size.");
    return 1;
  }
  unsigned pos = 0;
  printf("Enter data:\n");
  while (pos < size) {
    printf("%u = ", pos);
    scanf("%d", &nums[pos++]);
  }
  printf("The stored data is as follows:");
  while (pos)
    printf(" %d", nums[--pos]);
  free(nums);
  getchar();
  getchar();
  return 0;
}
```