Re: SFINAE

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Thu, 27 Mar 2008 22:46:06 -0400
Message-ID:
<wu6dnRhQU9JxxnHanZ2dnUVZ_vKunZ2d@comcast.com>
none" <""johs\"@(none) wrote:

I am trying to understand SFINAE based on this wiki page:

http://en.wikipedia.org/wiki/SFINAE

// Defines return type to be void for all types
// besides int.
template<typename T>
struct can_use_f {
typedef void type;
};

template<>
struct can_use_f<int> {

 typedef int type;
};

// function with return type ::type.
template<typename T>
typename can_use_f<T>::type f(T const &);

int main() {

f(1);
f(1.);

}

I have made a slight modification so if f(1) is called f should have
return type int. But it gives the error:

/tmp/cclJK0dJ.o: In function `main':
bob.cpp:(.text+0x95): undefined reference to `can_use_f<int>::type
f<int>(int const&)'
bob.cpp:(.text+0xa5): undefined reference to `can_use_f<double>::type
f<double>(double const&)'
collect2: ld returned 1 exit status


Of course it does. You declared the 'f' template function, but
you haven't given it a body!

Is this the correct execution steps:

1) f(T const &) is called.
2) the return type is decided based on the template parameter.
3) if T != int return type is void. If T = int return type should be
int (this does not work).


WHAT does not work? The Wikipedia article is not about that, is it?

V
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