c++ - C++ 将指针传递给期望 void* 的函数以两种不同的方式工作，为什么？

Question

最近我偶然发现了我融入这个例子的代码：

void PrintThis (void* input)
{
    std::cout << input << std::endl;
    std::cout << (char*)input << std::endl;
}

int main(void)
{
    char question[] = "Why does it work?";
    PrintThis(question);    // output: address + 'Why does it work?' - expected!
    PrintThis(&question);   // output: address + 'Why does it work?' - unexpected
    char* thisIsDiff = "This is different";
    PrintThis(thisIsDiff);    // output: address + 'This is different' - expected
    PrintThis(&thisIsDiff);   // output: address + rubbish - expected
    return 0;
}

请注意标有意外的行。为什么输出与上面的行相同？为什么当函数参数是 char* 而不是 void 时它甚至无法编译？

标签： c++functionvoidvoid-pointers