Question

Using GCC 4.2.
I have this metatemplate for conditional type:

template <bool condition, typename Then, typename Else>
struct IF
{
    typedef Then RET;
};

template <class Then, class Else>
struct IF<false, Then, Else>
{
    typedef Else RET;
};

and when I use it like this:

template <typename T>
class Param
{
    IF< sizeof(int)<sizeof(long), long, int>::RET  i;
};

it works, but when I use it like this (trying to use template parameters):

template <typename T>
class Param
{
    IF< sizeof(int)<sizeof(long), T&, T* >::RET mParam;
};

I am getting this error code:

error: type 'IF<false, T&, T*>' is not derived from type 'Param<T>'

Why is it happening? How to solve it?
Thanks in advance!

Answer 1

In the second case, what RET is, depends on the template type T. The compiler needs to be assured that it is going to be a type in all possible instantiations (and not perhaps a static member of some instantiation of IF). You do so with the typename keyword.

template <typename T>
class Param
{
    typename IF< sizeof(int)<sizeof(long), T&, T* >::RET mParam;

};