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
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  je     L79
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L35
L79:
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L48
L24:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  movsbl -0x5(%rbp),%eax
  sub    $0x41,%eax
  cmp    $0x38,%eax
  ja     L18
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D91(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D91(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D82(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
  cmpl   $0x2,-0x4(%rbp)
  jle    L59
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L59
L18:
  movsbl -0x5(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L91
L59:
  nop
L91:
  addl   $0x1,-0x4(%rbp)
L48:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L24
  mov    $0x0,%eax
L35:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D39
type_infered: string
name: disassemble_data
arguments: data_tag: D33
type_infered: string
name: disassemble_data
arguments: data_tag: D0
type_infered: string
name: disassemble_data
arguments: data_tag: D82
type_infered: string
name: disassemble_data
arguments: data_tag: D54
type_infered: string
name: disassemble_data
arguments: data_tag: D59
type_infered: string
name: disassemble_data
arguments: data_tag: D20
type_infered: string
name: disassemble_data
arguments: data_tag: D77
type_infered: string

tool: D39: ERROR: You need exactly one argument.

tool: D33: %d: 'A'


tool: D0: %d: 'E'


tool: D82: %d: 'I'


tool: D54: %d: 'O'


tool: D59: %d: 'U'


tool: D20: %d: 'Y'


tool: D77: %d: %c is not a vowel


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("ERROR: You need exactly one argument.\n");
    return 1;
  }
  int i = 0;
  for (i = 0; argv[1][i] != '\0'; i++) {
    char letter = argv[1][i];
    switch (letter) {
    case 'a':
    case 'A':
      printf("%d: 'A'\n", i);
      break;
    case 'e':
    case 'E':
      printf("%d: 'E'\n", i);
      break;
    case 'i':
    case 'I':
      printf("%d: 'I'\n", i);
      break;
    case 'o':
    case 'O':
      printf("%d: 'O'\n", i);
      break;
    case 'u':
    case 'U':
      printf("%d: 'U'\n", i);
      break;
    case 'y':
    case 'Y':
      if (i > 2) {
        printf("%d: 'Y'\n", i);
      }
      break;
    default:
      printf("%d: %c is not a vowel\n", i, letter);
      break;
    }
  }
  return 0;
}
```