user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x84(%rbp)
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x8c(%rbp)
  jmp    L22
L62:
  lea    -0x80(%rbp),%rdx
  mov    -0x8c(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x8c(%rbp)
L22:
  mov    -0x90(%rbp),%eax
  cmp    %eax,-0x8c(%rbp)
  jl     L62
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x8c(%rbp)
  jmp    L27
L119:
  lea    -0x40(%rbp),%rdx
  mov    -0x8c(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x8c(%rbp)
L27:
  mov    -0x90(%rbp),%eax
  cmp    %eax,-0x8c(%rbp)
  jl     L119
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x8c(%rbp)
  jmp    L10
L104:
  mov    -0x8c(%rbp),%eax
  cltq
  movzbl -0x80(%rbp,%rax,1),%edx
  mov    -0x8c(%rbp),%eax
  cltq
  movzbl -0x40(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L100
  mov    -0x8c(%rbp),%eax
  mov    %eax,-0x88(%rbp)
  jmp    L111
L12:
  mov    -0x88(%rbp),%eax
  cltq
  movzbl -0x80(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x88(%rbp)
L111:
  mov    -0x88(%rbp),%eax
  cltq
  movzbl -0x80(%rbp,%rax,1),%edx
  mov    -0x88(%rbp),%eax
  cltq
  movzbl -0x40(%rbp,%rax,1),%eax
  cmp    %al,%dl
  je     L23
  mov    -0x90(%rbp),%eax
  cmp    %eax,-0x88(%rbp)
  jl     L12
L23:
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L6
L31:
  mov    -0x8c(%rbp),%eax
  cltq
  movzbl -0x40(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8c(%rbp)
L6:
  mov    -0x8c(%rbp),%eax
  cmp    -0x88(%rbp),%eax
  jl     L31
  mov    $0xa,%edi
  call   <putchar@plt>
L100:
  addl   $0x1,-0x8c(%rbp)
L10:
  mov    -0x90(%rbp),%eax
  cmp    %eax,-0x8c(%rbp)
  jl     L104
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D26
data_type: string
name: unpack_data
arguments: data_tag: D40
data_type: string
name: unpack_data
arguments: data_tag: D65
data_type: string
name: unpack_data
arguments: data_tag: D38
data_type: string
name: unpack_data
arguments: data_tag: D30
data_type: string

tool: D16: Array  size : 

tool: D26: %d

tool: D40: Elements to array :

tool: D65:  %c

tool: D38: ----Output----

tool: D30:  , 

assistant: ```c
int main(void) {
  char a[50], b[50];
  int i, j, n, st = 0;
  printf("Array  size : ");
  scanf("%d", &n);
  printf("Elements to array :");
  for (i = 0; i < n; i++)
    scanf(" %c", &a[i]);
  printf("Elements to array :");
  for (i = 0; i < n; i++)
    scanf(" %c", &b[i]);
  printf("----Output----\n");
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      for (j = i; a[j] != b[j] && j < n; j++)
        printf("%c", a[j]);
      printf(" , ");
      for (; i < j; i++)
        printf("%c", b[i]);
      printf("\n");
    }
  }
  return 0;
}
```