I hope the subject didn't make you run away ;-) Let me be clearer (I'll try at least): If I have a template base class with a template method : template <typename T> class S { public: template <typename U> void f(U p, typename enable_if<is_same<T, U> >::type*dummy = 0) { std::cout << p << std::endl; } }; For the example, I simplify the method : it must "exists" only if T == U If A is this class: class A : public S<int> {}; Then I have what I want: int i = 1; A a; a.f(i); compiles, but double d = 2.0; a.f(d); doesn't compile : error: no matching function for call to âA::f(double&)â It is the expected behavior. Now let's A inherit from S<double> also : class A : public S<int>, public S<double> {}; Then the following code doesn't compile : int i = 1; A a; a.f(i); error: request for member âfâ is ambiguous error: candidates are: template<class U> void S::f(U, typename boost::enable_if<boost::is_same<T, U>, void>::type*) [with U = U, T = double] error: template<class U> void S::f(U, typename boost::enable_if<boost::is_same<T, U>, void>::type*) [with U = U, T = int] I expected there is no ambiguity : f<int> exists only for S<int> In the compiler error, we can notice that T is known when this piece of code is compiled, but not U (U = U). Any explanation or "workaround" ? Thanks. Nicolas.