user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x4c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  lea    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x38(%rbp)
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x48(%rbp)
  jmp    L79
L59:
  mov    -0x48(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x48(%rbp)
L79:
  mov    -0x4c(%rbp),%eax
  cmp    %eax,-0x48(%rbp)
  jl     L59
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x48(%rbp)
  jmp    L9
L40:
  mov    -0x48(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x48(%rbp)
L9:
  mov    -0x4c(%rbp),%eax
  cmp    %eax,-0x48(%rbp)
  jl     L40
  jmp    L85
L77:
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x44(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    -0x44(%rbp),%edx
  mov    %edx,(%rax)
  addq   $0x4,-0x40(%rbp)
  subq   $0x4,-0x38(%rbp)
L85:
  mov    -0x40(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L77
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x48(%rbp)
  jmp    L32
L53:
  mov    -0x48(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x48(%rbp)
L32:
  mov    -0x4c(%rbp),%eax
  cmp    %eax,-0x48(%rbp)
  jl     L53
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D65
type_infered: string
name: retrieve_data
arguments: label: D57
type_infered: string
name: retrieve_data
arguments: label: D54
type_infered: string
name: retrieve_data
arguments: label: D1
type_infered: string
name: retrieve_data
arguments: label: D4
type_infered: string
name: retrieve_data
arguments: label: D3
type_infered: string

tool: D65: 
Enter the no of elements in the 1st array

tool: D57: %d

tool: D54: 
Enter elements of the Array

tool: D1: 
Elements of the Array before reversing

tool: D4: %d	

tool: D3: 
Elements of the Array after reversing

assistant: ```c
int main() {
  int iaArr[10], iLen, i, iTemp;
  int *iptr1, *iptr2;
  printf("\nEnter the no of elements in the 1st array\n");
  scanf("%d", &iLen);
  iptr1 = iaArr;
  iptr2 = (iaArr + iLen - 1);
  printf("\nEnter elements of the Array\n");
  for (i = 0; i < iLen; i++)
    scanf("%d", (iptr1 + i));
  printf("\nElements of the Array before reversing\n");
  for (i = 0; i < iLen; i++)
    printf("%d\t", *(iptr1 + i));
  while (iptr1 < iptr2) {
    iTemp = *iptr1;
    *iptr1 = *iptr2;
    *iptr2 = iTemp;
    iptr1++;
    iptr2--;
  }
  iptr1 = iaArr;
  printf("\nElements of the Array after reversing\n");
  for (i = 0; i < iLen; i++)
    printf("%d\t", *(iptr1 + i));
  return 0;
}
```