user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0xc(%rbp)
  jmp    L7
L17:
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D85(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D97(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D85(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  cmp    $0x4,%eax
  ja     L4
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D112(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D112(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <no_of_digits@plt>
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D92(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <reverse@plt>
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse_original@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
  lea    D126(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x2,-0xc(%rbp)
  jmp    L7
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <isPalindrome@plt>
  jmp    L7
L4:
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
L7:
  cmpl   $0x1,-0xc(%rbp)
  je     L17
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D28
type_infered: string
name: parse_data
arguments: data_tag: D85
type_infered: string
name: parse_data
arguments: data_tag: D97
type_infered: string
name: parse_data
arguments: data_tag: D12
type_infered: string
name: parse_data
arguments: data_tag: D92
type_infered: string
name: parse_data
arguments: data_tag: D53
type_infered: string
name: parse_data
arguments: data_tag: D22
type_infered: string
name: parse_data
arguments: data_tag: D20
type_infered: string
name: parse_data
arguments: data_tag: D126
type_infered: string
name: parse_data
arguments: data_tag: D47
type_infered: string

tool: D28: 
Enter a number:

tool: D85: %d

tool: D97: Enter 1 to print the no of digits,2 to reverse the no using pass by value,3 to reverse no using pass by reference and 4 to check if entered no is pallindrome or not

tool: D12: Enter a menu indice:

tool: D92: The no of digits in %d is %d 


tool: D53: The original no is: %d 


tool: D22: The reversed no is %d

tool: D20: The original no is %d 


tool: D126: Program terminates

tool: D47: Im sorry I dont understand you:

assistant: ```c
int main() {
  int choice, no;
  int i = 1;
  while (i == 1) {
    puts("\nEnter a number:");
    scanf("%d", &no);
    puts("Enter 1 to print the no of digits,2 to reverse the no using pass by value,3 to reverse no using pass by reference and 4 to check if entered no is pallindrome or not");
    puts("Enter a menu indice:");
    scanf("%d", &choice);
    switch (choice) {
    case 1:
      printf("The no of digits in %d is %d \n", no, no_of_digits(no));
      break;
    case 2: {
      printf("The original no is: %d \n", no);
      printf("The reversed no is %d", reverse(no));
      break;
    }
    case 3: {
      printf("The original no is %d \n", no);
      reverse_original(&no);
      printf("The reversed no is %d", no);
      break;
    }
    case 0: {
      puts("Program terminates");
      i = 2;
      break;
    }
    case 4:
      isPalindrome(no);
      break;
    default:
      puts("Im sorry I dont understand you:");
      break;
    }
  }
  return 0;
}
```