Coversion operator template specialisation

This is a discussion on Coversion operator template specialisation within the C++ Programming forums, part of the General Programming Boards category; The following won't compile. If I comment out the conversion operator specialisation it compiles and runs without any problem. As ...

  1. #1
    DL1
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    Coversion operator template specialisation

    The following won't compile. If I comment out the conversion operator specialisation it compiles and runs without any problem. As far as I can see I have followed the normal syntax for template function specialisation. Presumably the rules for conversion operators are different. Can anybody point me in the right direction? Thanks.

    Code:
    #include <iostream>
    using namespace std;
    
    template <class T> class I1
    
    {
        public:
        T i, j;
        I1(T a = 0, T b = 0): i(a), j(b){}
        ~I1(){}
    
        template <class U> operator I1<U>()
        {
            return I1 <U> (U(i), U(j));
        }
    
    //problem with the following specialisation://
        template <> operator I1<char>()
        {
            cout << "char specialisation" << endl;
            return I1 <char> (char(i), char(j));
        }
    
    };
    
    
    int main()
    {
        I1 <int> var2(97, 99);
        cout << var2.i << ", " << var2.j << endl;
        I1 <char> var3(var2);
        cout << var3.i << ", " << var3.j << endl;
        return 0;
    }

  2. #2
    The larch
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    What if you just omit template<> since you normally just overload functions and don't specialize them.

    Also, from the use case, are you sure that you want to overload conversion operators. Wouldn't a converting constructor do?

    And I probably wouldn't use C-style casts in template context. static_cast<> will reject some invalid casts which a C-style cast might silently let pass.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  3. #3
    DL1
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    Thanks, overloading the function seems to solve the problem.

    However, my main aim was not to write an elegant program (and I am aware of the deficiencies of the program as it is written); I am just fiddling about trying to learn what you can and cannot do.

    So the question remains: is such a specialisation legal in principle, and if so, what is wrong with it as it stands?

  4. #4
    The larch
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    It appears that you can't specialize templated methods. You can specialize free functions, but things may not be as it would seem in that case. (Why not specialize function templates?)

    I can't remember ever having to specialize a function template and I think specialization is more useful for class templates.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  5. #5
    DL1
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    Why can you not specialise templated methods (I presume that "method" is the same as member function)? Section 14.5 of the draft standard gives the following example:

    Code:
    struct A {
        template <class T> operator T*();
    };
    template <class T> A::operator T*(){ return 0; }
    template <> A::operator char*(){ return 0; } // specialization
    template A::operator void*(); // explicit instantiation
    
    int main() {
        A a;
        int *ip;
        ip = a.operator int*(); // explicit call to template operator
                                // A::operator int*()
    }

  6. #6
    The larch
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    Well, I guess you can then, as long as you do it outside the class declaration.

    Since you've found the draft of the standard you can probably get better answers there.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  7. #7
    DL1
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    Unfortunately I can't find an answer in the standard, and even if I put the specialisation outside the class definition, like this:


    Code:
    template <class T>
        template <>
            I1<T>::operator I1<char>()
            {
                cout << "char specialisation" << endl;
                return I1 <char> (char(i), char(j));
            }


    I still can't get it to compile. I know it really is a trivial problem, but it's still driving me mad!

  8. #8
    Master Apprentice phantomotap's Avatar
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    You can't do it because the enclosing class isn't specialized. You can't partially or fully specialize a nested function or class unless the containing class is fully specialized. It is unfortunate, but the logic for this is sound. Do not even try it. You will only waste time. It can't work. (A compiler that does allow it is broken.)

    You can, usually, simply overload the function. I would suggest you always mark such a function as 'inline' so the source can stay in the header with the source for the template. (You may get linker errors if you don't.) That isn't mandatory.

    You can always specialize the interface and the mechanic of a nested function or class. The method, the technique, is ugly. It is also occasionally necessary if you employ a lot of sorcery. Put the technique in your bag of tricks for those occasions, but do try not to overuse it.

    This is ugly. I'm kind of busy. It will show you some of what I'm talking about.

    Soma

    Code:
    #include <iostream>
    using namespace std;
    
    template <class T> class I1;
    
    template <class T, class U> struct helper
    {
       static I1<U> go(I1<T> & f);
    };
    
    template <class T> class I1
    {
        public:
        T i, j;
        I1(T a = 0, T b = 0): i(a), j(b){}
        ~I1(){}
    
        template <class U> operator I1<U>()
        {
            return helper<T, U>::go(*this);
        }
    
    };
    
    template <class T, class U> I1<U> helper<T, U>::go(I1<T> & f)
    {
       return I1 <U> (U(f.i), U(f.j));
    }
    
    template <class T> struct helper<T, char>
    {
      static I1<char> go(I1<T> & f);
    };
    
    template <class T> I1<char> helper<T, char>::go(I1<T> & f)
    {
       cout << "char specialisation" << endl;
        return I1 <char> (char(f.i), char(f.j));
    }
    
    
    int main()
    {
        I1 <int> var2(97, 99);
        cout << var2.i << ", " << var2.j << endl;
        I1 <char> var3(var2);
        cout << var3.i << ", " << var3.j << endl;
        return 0;
    }

  9. #9
    DL1
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    Many, many thanks. I will have to give your answer and example program some thought. It looks rather complex right now. But that's the great thing about learning C++: every day you come up against things which seem almost impossible to get your head round, and then a day or two later they become perfectly clear. At least that's been my experience to date.

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