请教下 C++类型转换的问题 - V2EX
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请教下 C++类型转换的问题

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  •   demos May 15, 2020 2243 views
    This topic created in 2188 days ago, the information mentioned may be changed or developed.
    int main(int argc, char* argv[])
    {
    uint64_t num64 = 0x1122334455667788;
    char* p1 = 0;
    char* p2 = 0;
    int n1 = 0;
    int n2 = 0;

    printf("%d %d\r\n", sizeof(num64 + p1 - p2), sizeof(num64 + n1 - n2));
    return 0;
    }

    我在 vs2019 上编译成 Win32/Release 运行这段代码,会输出 4 8
    看起来笫一个运算结果被转换成了 int 型,而笫二个运算是 uint64_t 型
    理论上来说一个长类型(uint64_t)和一个短类型(int)的运算结果是长类型,结果应该两个都是 8 。
    但是笫一个结果是 4,想了很久没算明白,请大佬们指点下谢谢。
    knewbie
        1
    knewbie  
       May 15, 2020
    第一个是指针类型:win32 平台,指针长度是 4, 第二个输出在计算时会将 int 提升为 uint64 类型的,最终结果还是 uint64,所以 sizeof 的大小还是 uint64 类型的大小,即是 8
    demos
        2
    demos  
    OP
       May 15, 2020
    我傻了
    printf("%d %d\r\n", sizeof(num64 + (p1 - p2)), sizeof(num64 + (n1 - n2)));
    这样就是 8,8 了
    谢谢老哥
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