C 位明智不是运算符更改值类型？

Question

我正在测试一些 C 代码，运行遇到了一个错误，让我陷入了循环。我最终创建了以下代码作为测试，结果让我感到惊讶。

似乎 ~ 运算符以某种方式改变了值的解释（即它的类型），因为除非类型强制转换，否则测试失败。对正在发生的事情的任何见解表示赞赏。

代码：

int main()
{
    unsigned short one=1, two=1, three=1;
    one=~one;
    printf("One: %hu Two: %hu\n",one,two);
    printf("Test: %s\n",(one==~three)? "Pass": "Fail");
    printf("Test: %s\n",(one==(unsigned short)~three)? "Pass": "Fail");
    printf("Test: %s\n",(one==~(unsigned short)three)? "Pass": "Fail");
    
    return 0;
}

输出：

One: 65534 Two: 1
Test: Fail
Test: Pass
Test: Fail

Answer 1

unsigned int one = 1;
~one; // This expression has type int. So it's likely to be 0xFFFFFFFE

C 位明智不是运算符更改值类型？

C bit wise not operator changing values type?

c

logical-operators