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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <__globinit_function@plt>
  mov    $0x2,%esi
  mov    $0x9,%edi
  call   <__CrestCall@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <__CrestShort@plt>
  movzwl -0x10(%rbp),%eax
  movswq %ax,%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <__CrestLoad@plt>
  movzwl -0x10(%rbp),%eax
  cwtl
  mov    %eax,%edi
  call   <dbl@plt>
  mov    %ax,-0xe(%rbp)
  movzwl -0xe(%rbp),%eax
  movswq %ax,%rax
  mov    %rax,%rsi
  mov    $0xc,%edi
  call   <__CrestHandleReturn@plt>
  lea    -0xe(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xb,%edi
  call   <__CrestStore@plt>
  movzwl -0xe(%rbp),%eax
  movswq %ax,%rdx
  lea    -0xe(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x11,%edi
  call   <__CrestLoad@plt>
  mov    $0x3,%edx
  mov    $0x0,%esi
  mov    $0x10,%edi
  call   <__CrestLoad@plt>
  movzwl -0xe(%rbp),%eax
  cwtl
  add    $0x3,%eax
  cltq
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0xf,%edi
  call   <__CrestApply2@plt>
  mov    $0x9,%edx
  mov    $0x0,%esi
  mov    $0xe,%edi
  call   <__CrestLoad@plt>
  movzwl -0xe(%rbp),%eax
  cmp    $0x6,%ax
  sete   %al
  movzbl %al,%eax
  mov    %rax,%rdx
  mov    $0xc,%esi
  mov    $0xd,%edi
  call   <__CrestApply2@plt>
  movzwl -0xe(%rbp),%eax
  cmp    $0x6,%ax
  jne    L29
  mov    $0x1,%edx
  mov    $0x5,%esi
  mov    $0x12,%edi
  call   <__CrestBranch@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x14,%edi
  call   <__CrestLoad@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x15,%edi
  call   <__CrestStore@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L7
L29:
  mov    $0x0,%edx
  mov    $0x7,%esi
  mov    $0x13,%edi
  call   <__CrestBranch@plt>
  mov    $0x1,%edx
  mov    $0x0,%esi
  mov    $0x16,%edi
  call   <__CrestLoad@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x17,%edi
  call   <__CrestStore@plt>
  movl   $0x1,-0xc(%rbp)
  nop
L7:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x18,%edi
  call   <__CrestLoad@plt>
  mov    $0x19,%edi
  call   <__CrestReturn@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int main(void) {
  short a;
  short tmp;
  int __retres3;
  {
    __globinit_function();
    __CrestCall(9, 2);
    __CrestShort(&a);
    __CrestLoad(10, (unsigned long)(&a), (long long)a);
    tmp = dbl(a);
    __CrestHandleReturn(12, (long long)tmp);
    __CrestStore(11, (unsigned long)(&tmp));
    __CrestLoad(17, (unsigned long)(&tmp), (long long)tmp);
    __CrestLoad(16, (unsigned long)0, (long long)3);
    __CrestApply2(15, 0, (long long)((int)tmp + 3));
    __CrestLoad(14, (unsigned long)0, (long long)9);
    __CrestApply2(13, 12, (long long)((int)tmp + 3 == 9));
    if ((int)tmp + 3 == 9) {
      __CrestBranch(18, 5, 1);
      __CrestLoad(20, (unsigned long)0, (long long)0);
      __CrestStore(21, (unsigned long)(&__retres3));
      __retres3 = 0;
      goto return_label;
    } else {
      __CrestBranch(19, 7, 0);
      __CrestLoad(22, (unsigned long)0, (long long)1);
      __CrestStore(23, (unsigned long)(&__retres3));
      __retres3 = 1;
      goto return_label;
    }
  return_label : {
    __CrestLoad(24, (unsigned long)(&__retres3), (long long)__retres3);
    __CrestReturn(25);
    return (__retres3);
  }
  }
}
```