c++ - Reference_wrapper 不会在 cout 中打印，但 reference_wrapper 会吗？

Question

为什么我尝试打印“string的reference_wrapper”的行会给出不支持的运算符<<“for“string的reference_wrapper”的错误，但不会给出“reference_wrapper for int”的错误？

int main(){

    int  s= 43;
    string str = "hello";

    reference_wrapper<int> x{s};
    reference_wrapper<string> y{str};

    x.get() = 47;
    y.get() = "there";

    cout<<"printing original int "<<s<<"\n";
    cout<<"printing original string "<<str<<"\n";

    cout<<"printing reference_wrapper for int "<<x<<"\n";
    cout<<"printing reference_wrapper for string "<<y<<"\n"; // gives error

    int& refint = x;
    string& refstr = y;

    cout<<"printing reference for int "<<refint<<"\n";
    cout<<"printing reference for string "<<refstr<<"\n";
}

Answer 1

operator<< for std::string 是一个函数模板，当传递 reference_wrapper 时，最后一个模板参数 Allocator无法推论；因为 template argument deduction 中不会考虑隐式转换.

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

作为解决方法，您可以调用 std::reference_wrapper<T>::get 显式或执行显式转换。

另一方面， operator<< for int 是非模板，则不存在此类问题。