Question

I had a wrong perception that template function in a class is instantiated only if it’s invoked. See the below simple code:

template<typename T>
struct A
{
  T *p; 
  T& operator * () { return *p; }
};

int main ()
{
  A<int> ai;   // ok
  int i = *ai;  // works fine
  A<void> av;  // compiler complains even "*av" is not called
}

Just while declaring A<void>, compiler errors out as:

error: forming reference to void

I tried to specialize the function for void outside the template as below:

template<>
void A<void>::operator * () {}

But it doesn’t help and gives error as:

error: no member function ‘operator*’ declared in ‘A<void>’

Is there any way to fix this with C++03 ?

Answer 1

What about

template < typename T >
struct ReferenceOrVoid
{ typedef T& Value; };

template < >
struct ReferenceOrVoid < void >
{ typedef void Value; };

template<typename T>
struct A
{
    T *p; 
    typename ReferenceOrVoid < T > :: Value
    operator * () { return *p; }
};

Of course it depends what you want A to behave in case T is void. You can also specialise the whole A struct for void, of course.