user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    -0x3c(%rbp),%r12
  push   %rbx
  lea    D5(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  nop
L14:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  lea    -0x3(%rsi),%eax
  cmp    $0x6,%eax
  ja     L14
  movslq %esi,%rax
  mov    %rsp,%rcx
  mov    %rax,-0x50(%rbp)
  imul   %rax,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L70
L52:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L52
L70:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L88
L97:
  xor    %eax,%eax
  mov    %esi,%edx
  lea    D43(%rip),%rdi
  mov    %rsp,-0x48(%rbp)
  call   <printf@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L113
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  lea    D66(%rip),%rbx
  nop
L21:
  mov    -0x48(%rbp),%rax
  xor    %r15d,%r15d
  lea    (%rax,%r13,4),%r14
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  add    $0x1,%r15d
  add    $0x4,%r14
  call   <scanf@plt>
  cmp    %r15d,-0x3c(%rbp)
  jg     L40
  mov    $0xa,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  add    -0x50(%rbp),%r13
  cmp    %r12d,-0x3c(%rbp)
  jg     L21
L113:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    -0x3c(%rbp),%esi
  test   %esi,%esi
  jle    L49
  mov    -0x50(%rbp),%r15
  mov    -0x48(%rbp),%rbx
  xor    %r13d,%r13d
  lea    D63(%rip),%r12
  shl    $0x2,%r15
  nopw   0x0(%rax,%rax,1)
L82:
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L0:
  mov    (%rbx,%r14,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <printf@plt>
  cmp    %r14d,-0x3c(%rbp)
  jg     L0
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    %r15,%rbx
  call   <putchar@plt>
  mov    -0x3c(%rbp),%esi
  cmp    %r13d,%esi
  jg     L82
L49:
  mov    -0x48(%rbp),%rdi
  call   <isMagic@plt>
  test   %eax,%eax
  jne    L129
  lea    D15(%rip),%rdi
  call   <puts@plt>
L62:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L129:
  lea    D56(%rip),%rdi
  call   <puts@plt>
  jmp    L62
L88:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L97
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D5
type: string
name: read_data
arguments: label: D43
type: string
name: read_data
arguments: label: D66
type: string
name: read_data
arguments: label: D34
type: string
name: read_data
arguments: label: D63
type: string
name: read_data
arguments: label: D15
type: string
name: read_data
arguments: label: D56
type: string

tool: D5: %d

tool: D43: Farewell!Now enter the values of you own %d by %d square:


tool: D66: %d%*c

tool: D34: Your "magic square" should look like this:

tool: D63: 	%d 

tool: D15: Your square indeed is magic!

tool: D56: Sorry but your square isn't magic!

assistant: ```c
int main() {
  int n, i = 0, j;
  do
    scanf("%d", &n);
  while (n <= 2 || n >= 10);
  int square[n][n];
  printf("Farewell!Now enter the values of you own %d by %d square:\n", n, n);
  for (; i < n; i++) {
    for (j = 0; j < n; j++) {
      scanf("%d%*c", &square[i][j]);
    }
    printf("\n");
  }
  printf("Your \"magic square\" should look like this:\n");
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      printf("\t%d ", square[i][j]);
    printf("\n");
  }
  if (!isMagic((int **)square, n))
    printf("Your square indeed is magic!\n");
  else
    printf("Sorry but your square isn't magic!\n");
  return 0;
}
```