关于c ++：operator＆lt;＆lt;（ostream＆amp;，const BigUnsigned＆lt; I＆gt;＆amp;）必须只接受一个参数


operator<<(ostream&, const BigUnsigned<I>&) must take exactly one argument

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

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template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector _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: warning: friend declaration
'std::ostream& operator<<(std::ostream&, const BigUnsigned&)' declares a non-template function [-Wnon-template-friend] friend std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu); ^ ../hw06/bigunsigned.h:13:77: note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here) ../hw06/bigunsigned.h:16:51:
error: invalid use of template-name 'BigUnsigned' without an argument
list std::ostream& operator<<(std::ostream& out, const BigUnsigned& bu){ ^ ../hw06/bigunsigned.h: In function 'std::ostream& operator<<(std::ostream&, const int&)': ../hw06/bigunsigned.h:17:28: error: request for member '_integers' in 'bu', which is of non-class type 'const int' for (auto integer : bu._integers){ ^

号

当我加入这样的声明和定义时，一切都很好。

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template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector _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;
    }
};

号

目的是将成员变量_integer打印到cout。可能有什么问题？

附言：使用这个问题，我使函数免费，但没有帮助。

相关讨论

BigUnsigned是模板类型，因此

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

不工作，因为没有BigUnsigned。你需要让friend函数成为一个模板，这样你就可以使用不同类型的BigUnsigned。

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template <typename I>
class BigUnsigned{
    const size_t cell_size=sizeof(I);
    std::vector _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;
}

号

第二个示例之所以有效，是因为它是在类内声明的，所以它使用类使用的模板类型。

相关讨论

纳撒诺利弗对答案的提炼。

对于另一个答案，函数模板的所有实例化都是类模板所有实例的friend。

operator<< 是BigUnsigned和BigUnsigned的friend。

operator<< 是BigUnsigned和BigUnsigned的friend。

您可以稍微更改声明，以便

operator<< 是BigUnsigned的friend而不是BigUnsigned的friend。

operator<< 是BigUnsigned的friend而不是BigUnsigned。

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// 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& bu);

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

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

要添加第三个稍微提高可读性的变量，需要在类内定义friend函数：

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#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 << '
';
}

。