Re: Class template specialization with template parameter
First, thanks for the reply.
Foo<int> uses the normal template, but Foo<std::map<T1, T2> > for any
types of T1 and T2 uses the specifalization?
Not sure what you're asking here.
What I was trying to say was that if I declare:
The instantiation will be for the normal/not-specialized template.
Yes. Of course, 'myFoo' in that case is a function, but the template
is still instantiated, I believe.
But if I declare:
Foo<std::map<int, std::string> > myFoo2();
Same thing. It's a function. But the 'std::map'-based specialisation
should be used, of course.
it will use the specialization... and that regardless of the types
used for std::map (in this case, int and std::string), any
"Foo<std::map<typeA, typeB> > myFooExample()" will always
instantiate the specialization.
Yes. Or do you still have a doubt about that?
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