CODEWITHSUNDEEP
c++templates
Sut
Sut

Reputation: 87

Forward declaration of C++ template

I want to add some friend declarations to a class. For example, I want to add some functions of operator== ,operator <. So what I have to do is use the forward declarations:

template <typename >
class MyBlob;

template <typename T>
bool operator==(const MyBlob<T> &, const MyBlob<T>&);

template <typename T>
bool operator!=(const MyBlob<T> &, const MyBlob<T>&);


template <typename T>
bool operator<(const MyBlob<T> &, const MyBlob<T>&);


template <typename T>
class MyBlob
{

    friend bool operator== <T>(const MyBlob<T> &lhs,const MyBlob<T> &rhs);
    friend bool operator!= <T>(const MyBlob<T> &lhs,const MyBlob<T> &rhs);
    friend bool operator< <T>(const MyBlob<T> &lhs,const MyBlob<T> &rhs);
      //other things

};

This is annoying that I have to use template <typename T> for three times. And this really reduce the readability.

So, is there any method to make the forward declaration more simple? Or can I have some method to declare these things in one place just like the ordinary function?
If this can't be done, Is using typedef to simplify the template <typename T> a good idea?

Upvotes: 0

Views: 92

Answers (1)

Grisha
Grisha

Reputation: 536

You can try defining friend operators within the class declaration:

template <typename T>
class MyBlob {

    friend bool operator== (const MyBlob& lhs, const MyBlob& rhs) {
        // ...
    }

    friend bool operator!= (const MyBlob& lhs, const MyBlob& rhs) {
        // ...
    }

    friend bool operator< (const MyBlob& lhs, const MyBlob& rhs) {
        // ...
    }

    // ...
};

Upvotes: 1

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