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Operator << for inner class of template class

I want to implement operator <<

that will print the contents of the inner class of the template class i.e. X<T>::Y

... It is quite possible to add any ads friend

. How to do it?

Here is a complete example of what I want to compile in C ++ 11:

#include <iostream>

template <typename T>
class X {
public:
    X(T d): y(d) {}

private:
    class Y {
    public:
        Y(T d): data(d) {}

    private:
        T data;

        template <typename U>
        friend std::ostream &operator <<(std::ostream &os, const typename X<U>::Y &y);
    };

    Y y;

    template <typename U>
    friend std::ostream &operator <<(std::ostream &os, const X<U> &x);
};

// This operator is not detected by the operator below
template <typename U>
std::ostream &operator <<(std::ostream &os, const typename X<U>::Y &y) {
    os << "Y(" << y.data << ")";
    return os;
}

template <typename U>
std::ostream &operator <<(std::ostream &os, const X<U> &x) {
    os << "X " << x.y;
    return os;
}

int main() {
    std::cout << X<int>(1);
    return 0;
}

Compiler error:

error: no match for 'operator<<' (operand types are 'std::basic_ostream<char>' and 'const X<int>::Y')
 os << "X " << x.y;
            ^
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1 answer


This does not work because it typename X<U>::Y

is a dependent type. If you want to print eg. a int

if the compiler just checks all possible types for U

(there are infinitely many due to recursion) to check if the typename X<U>::Y

result is int

(consider specialization!)? Therefore, the only possibility is to implement the function directly in the friend statement:

class Y {
public:
    Y(T d): data(d) {}

private:
    T data;

    friend std::ostream &operator <<(std::ostream &os, Y const& y)
    {
        os << "Y(" << y.data << ")";
        return os;
    }
};


http://ideone.com/dx8Qri

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