为什么“无符号 int64_t"在 C 中给出错误?

问题描述

为什么下面的程序会报错?

Why the following program gives an error?

#include <stdio.h>

int main()
{
    unsigned int64_t i = 12;
    printf("%lld\n", i);
    return 0;
}

错误:

 In function 'main':
5:19: error: expected '=', ',', ';', 'asm' or '__attribute__' before 'i'
  unsigned int64_t i = 12;
                   ^
5:19: error: 'i' undeclared (first use in this function)
5:19: note: each undeclared identifier is reported only once for each function it appears in

但是，如果我删除 unsigned 关键字，它就可以正常工作.所以，为什么 unsigned int64_t i 会报错?

But, If I remove the unsigned keyword, it's working fine. So,Why unsigned int64_t i gives an error?

推荐答案

您不能在 int64_t 类型上应用 unsigned 修饰符.它只适用于 char、short、int、long 和 long long.

You cannot apply the unsigned modifier on the type int64_t. It only works on char, short, int, long, and long long.

您可能想要使用 uint64_t，它是 int64_t 的无符号对应物.

You probably want to use uint64_t which is the unsigned counterpart of int64_t.

另请注意，int64_t 等.在头文件 stdint.h 中定义，如果你想使用这些类型，你应该包括它.

Also note that int64_t et al. are defined in the header stdint.h, which you should include if you want to use these types.

这篇关于为什么“无符号 int64_t"在 C 中给出错误?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！