如何在C++中为模板重载朋友提取操作符(>>)？

Question

我试图>>用模板重载friend 操作符.我不想内联定义它.

我曾尝试add()在下面的代码中定义的方法的帮助下做同样的事情.它工作正常.我希望我的>>运营商也这样做.

以下是我的代码:

#include<iostream>

template<class T>class Demo;
template<class T>
std::ostream& operator<<(std::ostream&, const Demo<T> &);
template<class T>
std::istream& operator>>(std::istream&, const Demo<T> &);

template<class T>
class Demo {
private:
    T data; // To store the value.
public:
    Demo(); // Default Constructor.
    void add(T element); // To add a new element to the object.
    Demo<T> operator+(const Demo<T> foo);
    friend std::ostream& operator<< <T>(std::ostream &out, const Demo<T> &d);
    friend std::istream& operator>> <T>(std::istream &in, const Demo<T> &d);
};

template<class T>
Demo<T>::Demo() {
    data = 0;
}   

template<class T>
void Demo<T>::add(T element) {
    data = element;
}

template<class T>
Demo<T> Demo<T>::operator+(const Demo<T> foo) {
    Demo<T> returnObject;
    returnObject.data = this->data + foo.data;
    return returnObject;
}

template<class T>
std::ostream& operator<<(std::ostream &out, const Demo<T> &d) {
    out << d.data << std::endl;
    return out;
}

template<class T>
std::istream& operator>>(std::istream &in, const Demo<T> &d) {
    in >> d.data;
    return in;
}

int main() {
    Demo<int> objOne;
    std::cin>>objOne;
    Demo<int>objTwo;
    objTwo.add(3);
    Demo<int>objThree = objOne + objTwo;
    std::cout << "Result = " << objThree;
    return 0;
}

实际问题

在尝试重载友元提取operator(>>)时,编译器会显示错误,如下所示:

testMain.cpp:52:15:   required from here
testMain.cpp:46:8: error: no match for 'operator>>' (operand types are 'std::istream {aka std::basic_istream}' and 'const int')
     in >> d.data;
        ^

预期产出

Result = 59

RUN SUCCESSFUL (total time: 49ms)

参考

• "在C++中重载提取运算符>> [duplicate]"定义如何重载>>运算符.但它没有提到它与模板的用法.

• "为模板类重载friend operator <<"定义了如何重载<<运算符,但不是>>.

• "重载提取运算符"也未涵盖模板概念.

Answer 1

该问题与模板无关.

operator>>修改右侧的数据,但是您将该参数声明为const,因此操作员无法修改它.编译器错误甚至声明要修改的值(d.data)是const:

testMain.cpp:46:8:错误:'运算符>>不匹配(操作数类型是'std :: istream {aka std :: basic_istream}'和'const int')

您需要const从第二个参数中删除:

template<class T>
std::istream& operator>>(std::istream&, Demo<T> &);

...

template<class T>
class Demo {
   ...
public:
    ...
    friend std::istream& operator>> <T>(std::istream &in, Demo<T> &d);
};

...

template<class T>
std::istream& operator>>(std::istream &in, Demo<T> &d) {
    in >> d.data;
    return in;
}