Reputation: 3329
I have a scenario in which there is a template class
template<typename X, typename Y>
class Foo
{
typedef Y::NestedType Bar;
int A (Bar thing);
void B();
int C(X that);
// other stuff
};
and then I would like the A() method to have a different behavior when X is a given type (but B and C can stay the same, and the actual code actually has about 10 other methods, a few of which are quite lengthy and subject to frequent tweaking.. so I would rather avoid making a full-class specialization and copy&paste the full class implementation)
I went on and wrote:
template<typename T>
int Foo<MyType, T>::A(Bar thing);
but my compiler (clang 163.7.1) refused to even consider this as a template specialization of any sort.
Is there some syntax error hidden in the way I wrote the code, or is this coding style invalid C++? Unfortunately, even if other compilers do support the idiom, my company is stuck with clang.
Thanks for any help on this.
Upvotes: 7
Views: 2206
Reputation: 506837
Use overloading
template<typename X, typename Y>
class Foo
{
// allows to wrap up the arguments
template<typename, typename>
struct Types { };
typedef Y::NestedType Bar;
int A (Bar thing) {
return AImpl(thing, Types<X, Y>());
}
void B();
int C(X that);
// other stuff
private:
template<typename X1, typename Y1>
int AImpl(Bar thing, Types<X1, Y1>) {
/* generic */
}
template<typename Y1>
int AImpl(Bar thing, Types<SpecificType, Y1>) {
/* special */
}
};
You cannot partially specialize a member of a class template. What you wrote would be the definition of a member function A
of a partial specialization of the class template itself.
Upvotes: 7