运算符

Question

我试图将我的模板化类的模板化成员函数的声明和定义分开，但最终出现了以下错误和警告。

template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector<I> _integers;
public:
    BigUnsigned();
    BigUnsigned(I);
    friend std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu);
};

std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu){
    for (auto integer : bu._integers){
        out<<integer<<std::endl;
    }
    return out;
}

../HW06/BigUnsigned.h:13:77:警告：友元声明“std::oStream&operator<<(std::oStream&,const BigUnsigned&)”声明了一个非模板函数[-wnon-template-friend]友元std::oStream&operator<<(std::oStream&out,const BigUnsigned&bu)；^../Hw06/BigUnsigned.h:13:77:注意：（如果这不是您想要的，请确保函数模板已经声明，并在此处的函数名后面添加<>)../Hw06/BigUnsigned.h:16:51:错误：在没有参数列表的情况下使用模板名“BigUnsigned”无效<<(std::Ostream&out，const BigUnsigned&bu){^../Hw06/BigUnsigned.h:In函数“std::Ostream&out，const Int&)”：../Hw06/BigUnsigned.h:在“bu”中请求成员“_integers”，该成员是非类

当我加入这样的声明和定义时，一切都编译得很好。

template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector<I> _integers;
public:
    BigUnsigned();
    BigUnsigned(I);
    friend std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu){
        for (auto integer : bu._integers){
            out<<integer<<std::endl;
        }
        return out;
    }
};

目的是将成员variableintegers打印到cout。可能是什么问题？

附注：使用这个问题，我使函数免费，但没有帮助。

Answer 1

要添加第三个改进可读性的变体,请在类中定义friend函数：

#include <iostream>

template <typename T>
class Foo {
    int test = 42;

    // Note: 'Foo' inside the class body is basically a shortcut for 'Foo<T>'
    // Below line is identical to: friend std::ostream& operator<< (std::ostream &os, Foo<T> const &foo)
    friend std::ostream& operator<< (std::ostream &os, Foo const &foo) {
        return os << foo.test;
    }
};


int main () {
    Foo<int> foo;
    std::cout << foo << '\n';
}

Answer 2

NathanOliver对答案的改进。

对于另一个答案，函数模板的所有实例化都是类模板的所有实例化的friends。

运算符< 是bigunsigned 和bigunsigned 的朋友。

运算符< 是bigunsigned 以及bigunsigned 的友元。

您可以稍微更改声明，以便

运算符< 是bigunsigned 的朋友但不是bigunsigned 的。

运算符< 是bigunsigned 的友元但不是bigunsigned 。

// Forward declaration of the class template.
template <typename I> class BigUnsigned;

// Forward declaration of the function template
template <typename I>
std::ostream& operator<<(std::ostream& out, const BigUnsigned<I>& bu);

// Change the friend-ship declaration in the class template.
template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector<I> _integers;
public:
    BigUnsigned();
    BigUnsigned(I);

    // Grant friend-ship only to a specific instantiation of the
    // function template.
    friend std::ostream& operator<< <I>(std::ostream& out, const BigUnsigned<I>& bu);
};

Answer 3

bigunsigned是模板类型，因此

std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu)

不会工作，因为没有bigunsigned。您需要使friend函数成为一个模板，以便可以使用不同类型的bigunsigned 。

template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector<I> _integers;
public:
    BigUnsigned();
    BigUnsigned(I);
    template<typename T>
    friend std::ostream& operator<<(std::ostream& out, const BigUnsigned<T>& bu);
};

template<typename T>
std::ostream& operator<<(std::ostream& out, const BigUnsigned<T>& bu){
    for (auto integer : bu._integers){
        out<<integer<<std::endl;
    }
    return out;
}

第二个示例起作用的原因是，由于它是在类内部声明的，所以它使用了类使用的模板类型。