为什么显式调用operator << ambiguous？

Question

这是简单的代码:

int main()
{
    int x=0;
    std::cout<<x; 
    operator<<(std::cout,x); //ambiguous

    return 0;
}

为什么operator<<(std::cout,x)电话不明确但不是std::cout<<x;？谢谢

Answer 1

这里的问题是输出整数,operator<<是一个std::ostream 成员函数.

So to explicitly call the operator function you should do e.g.

std::cout.operator<<(x);

The stand-alone operator<< function is for characters and strings.