user: 这段汇编代码用C语言表示是什么样的:

```asm
<palindrome>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  je     L12
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %eax,%ecx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  mov    %eax,(%rsp)
  test   %ecx,%ecx
  je     L12
  movslq %ecx,%rdx
  mov    %ecx,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edi
  mov    %edx,%esi
  mov    %ecx,%edx
  add    %edi,%edi
  mov    %esi,%r8d
  sub    %edi,%edx
  mov    %edx,0x4(%rsp)
  test   %esi,%esi
  je     L150
  movslq %esi,%rcx
  mov    %esi,%edi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%edi
  sar    $0x22,%rcx
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%r8d
  mov    %ecx,%edi
  add    %r8d,%r8d
  sub    %r8d,%esi
  mov    %ecx,%r8d
  mov    %esi,0x8(%rsp)
  test   %ecx,%ecx
  je     L190
  movslq %ecx,%rcx
  mov    %edi,%esi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%esi
  sar    $0x22,%rcx
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%r8d
  mov    %ecx,%esi
  add    %r8d,%r8d
  sub    %r8d,%edi
  mov    %ecx,%r8d
  mov    %edi,0xc(%rsp)
  test   %ecx,%ecx
  je     L1
  movslq %ecx,%rcx
  mov    %esi,%edi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%edi
  sar    $0x22,%rcx
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%r8d
  mov    %ecx,%edi
  add    %r8d,%r8d
  sub    %r8d,%esi
  mov    %ecx,%r8d
  mov    %esi,0x10(%rsp)
  test   %ecx,%ecx
  je     L40
  movslq %ecx,%rcx
  mov    %edi,%esi
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%esi
  sar    $0x22,%rcx
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%esi
  add    %esi,%esi
  sub    %esi,%edi
  mov    %edi,0x14(%rsp)
  cmp    %eax,%edi
  jne    L195
  mov    $0x5,%ecx
L159:
  lea    -0x1(%rcx),%esi
  movslq %esi,%rsi
  cmp    %edx,(%rsp,%rsi,4)
  jne    L195
  lea    -0x2(%rcx),%edx
  mov    0x8(%rsp),%esi
  movslq %edx,%rdx
  cmp    %esi,(%rsp,%rdx,4)
  jne    L195
  lea    -0x3(%rcx),%edx
  mov    0xc(%rsp),%esi
  movslq %edx,%rdx
  cmp    %esi,(%rsp,%rdx,4)
  jne    L195
  cmp    $0x3,%ecx
  je     L12
  lea    -0x4(%rcx),%edx
  mov    0x10(%rsp),%edi
  movslq %edx,%rdx
  cmp    %edi,(%rsp,%rdx,4)
  jne    L195
  cmp    $0x5,%ecx
  jne    L12
  cmp    %eax,0x14(%rsp)
  jne    L195
L12:
  mov    $0x1,%r8d
L111:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  mov    %r8d,%eax
  add    $0x28,%rsp
  ret
L1:
  cmp    %edi,%eax
  jne    L111
  mov    $0x3,%ecx
  jmp    L159
  nopl   0x0(%rax)
L195:
  xor    %r8d,%r8d
  jmp    L111
  nopl   (%rax)
L150:
  cmp    %edx,%eax
  je     L12
  jmp    L111
  cs nopw 0x0(%rax,%rax,1)
L190:
  cmp    %esi,%eax
  je     L12
  jmp    L111
  cs nopw 0x0(%rax,%rax,1)
L40:
  cmp    %esi,%eax
  jne    L111
  mov    $0x4,%ecx
  jmp    L159
L86:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int palindrome(int num) {
  int digit = 0;
  int numarray[6];
  while (num != 0) {
    numarray[digit] = num % 10;
    num /= 10;
    digit++;
  }
  int numdigits = digit - 1;
  int revarray[numdigits];
  for (digit = 0; digit <= numdigits; digit++) {
    revarray[digit] = numarray[numdigits - digit];
    if (!(revarray[digit] == numarray[digit])) {
      return 0;
    }
  }
  return 1;
}
```