Re: Specifying the leaf type in templates
Coder Guy wrote:
The logic I am trying to accomplish is this:
class Test : public Base<"LeafType"> { };
For example, this works:
class Test : public Base<Test> { };
Is there anyway to have the leaf type be the current type if it is not
derived, but if it is derived it should be the derived type. For example:
template<class L = Test>
class Test : public Base<L> { };
This compiles fine until instantiate the template class.
For a start, Test isn't a class, it's a class template.
I get this error:
Test t; // error C2955: 'Test' : use of class template requires template
argument list
Right.
Any input is appreciated.
This may or may not meet your needs, depending on what you are trying to
achieve (you've only shown us how you're trying to achieve it):
template<class L = void>
class Test;
template <class L>
class Test: public Base<L>
{
};
template <>
class Test<void>: public Base<Test<void> >
{
};
That way, Test<void> is the non-derived version, and Test<Derived> is
the version that Derived derives from.
Tom
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