operator overloading with templates [duplicate]

Question

I have a class which must be able to hold a type of float, double, long excetera. I would like to overload it in such a way that it can add two instances holding different types.

template <typename T> class F{ public: F(const T & t){a=t;} T a; F & operator+= (const F & rhs); } template<typename T> F<T> F::operator+= (const F & rhs){ a+=rhs.a; return *this

This is just pseudo code I've kept out irrelevant bits where I'm actually trying to use this sort of solution.

Now when trying to use:

 F<int> first(5); F<int> second(4); first+=second; // Works F<int> third(5); F<float> fourth(6.2); fourth+=third; // wont work

I can see why this doesn't work as it assumes the rhs argument is the same type as the lhs. I can also see there are potential problems in performing an operation which is int += long as if the long is big the type would need changing. What I can't seem to find is a nice way to solve the problem. I would appreciate your inputs. Thanks

Seth Carnegie · Accepted Answer · 2011-12-12 21:46:36Z

7

You need to make operator+= a template as well:

template <typename T> class F{ public: F(const T & t){ a = t; } T a; template<typename T2> F& operator+=(const F<T2>& rhs); }; // semicolon was missing template<typename T> template<typename T2> F<T>& F<T>::operator+=(const F<T2>& rhs) { a += rhs.a; return *this; // semicolon was missing } // brace was missing

Then you can do

F<int> a(4); F<float> b(28.4); b += a; cout << b.a << endl; // prints 32.4

Here is a working example.

edited Dec 12, 2011 at 21:46

answered Dec 12, 2011 at 21:41

Seth Carnegie

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3 Comments

wookie1 Over a year ago

Thanks Seth, I had tried mokas answer first then your edit in my code but it wouldn't compile. Thanks to your answer I believe I know the reason now. Because my code is quite large I had divided it into header and implementation using the cheat of declaring the templates I would use in the implementation file, ie template class F<float>; and template class F<int>; but forgot to provide one for the function. Now I'm struggling to make one, I think it should be something like this: template Variable<double> Variable<double>::operator+=(const Variable<float> & other); but that doesn't work.

Aaron McDaid Over a year ago

@wookie1, I think you're missing a & in your comment. template Variable<double>& Variable<double>::operator+=(const Variable<float> & other);

wookie1 Over a year ago

Ah, silly me I was trying all kinds of combinations trying to get that working, thanks.

moka · Accepted Answer · 2011-12-12 21:46:14Z

 template <typename T> class F{ public: F(const T & t){a=t;} T a; template<typename T2> F & operator+= (const F<T2> & rhs); }; template<typename T> template<typename T2> F<T>& F<T>::operator+= (const F<T2> & rhs){ a+=(T)rhs.a; return *this }

EDIT:

fixed error, see comments

This is the right idea, but it won't compile; see @Seth's answer.

John Dibling · Accepted Answer · 2011-12-12 21:43:34Z

You can templatize operator+=:

template<typename G> F<T> & operator+= (const F<G> & rhs);

...assuming you can somehow convert a G to a T.

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