[boost] enable_if : how to make it work in the case of templated method of a template base inherited several times

Subject: [boost] enable_if : how to make it work in the case of templated method of a template base inherited several times
From: nico (nicolas.guillot_at_[hidden])
Date: 2011-10-20 10:32:39

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 ‘A::f(double&)’
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 ‘f’ 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.