Question

我不知道在尝试重载＆gt;＆gt;时出了什么问题和＆lt;＆lt;

template<class T, int N> class vector{
    friend istream &operator>> <T,N>(istream &, vector &);
    friend ostream &operator<< <T,N>(ostream &, const vector &);
};

template<class T, int N>
istream &operator>>(istream &input, vector<T,N>& v)
{
    for (int i=0; i<N; i++)
        input >> v.component[i];
    return input;
}

g ++编译器说

vector.h:22:17: error: template-id ‘operator>><double, 2>’ for 
‘std::istream& operator>>(std::istream&, vector<double, 2>&)’ 
does not match any template declaration
friend istream &operator>> <T,N>(istream &, vector &);
             ^
vector.h:23:17: error: template-id ‘operator<< <double, 2>’ for
‘std::ostream& operator<<(std::ostream&, const vector<double, 2>&)’
does not match any template declaration
friend ostream &operator<< <T,N>(ostream &, const vector &);
             ^

Answer 1

你需要在模板成为朋友之前声明模板（即，friend子句不能像为函数那样声明模板）。

template<class T, int N> class vector;
template<class T, int N>
istream &operator>>(istream &input, vector<T,N>& v);
template<class T, int N>
ostream &operator<<(ostream &input, vector<T,N> const& v);

template<class T, int N> class vector{
    friend istream &operator>> <T,N>(istream &, vector &);
    friend ostream &operator<< <T,N>(ostream &, const vector &);
};

template<class T, int N>
istream &operator>>(istream &input, vector<T,N>& v)
{
    for (int i=0; i<N; i++)
        input >> v.component[i];
    return input;
}

Answer 2

您编写重载的方式表示函数模板特化。因此错误，因为没有先前声明的主模板。如果这是主要定义（），则只需要封闭类声明中的模板参数，因为当您使用类名作为参数时，它们会被强制执行：

template<class T, int N> class vector {
    friend istream &operator>> (istream &, vector &);
    friend ostream &operator<< (ostream &, const vector &);
};

这仅适用于函数的类内声明。您还可以接受不同类型和值的模板参数：

template<class T, int N> class vector {
    template<class T2, int N2>
    friend istream &operator>> (istream &, vector<T2, N2> &);

    template<class T2, int N2>
    friend ostream &operator<< (ostream &, const vector<T2, N2> &);
};

模板和重载istream

2 个答案: