Question

I have a template function:

template<typename T>
void foo(const T& value) { bar(value); x = -1; }

I want to specialize it for a set of types:

template<>
void foo<char>(const char& value) { bar(value); x = 0; }
template<>
void foo<unsigned char>(const unsigned char& value) { bar(value); x = 1; }

It works ok. When I compile this:

template<>
void foo<char*>(const char*& value) { bar(value); x = 2; }

I get an error:

error C2912: explicit specialization; 'void foo(const char *&)' is not a specialization of a function template

Is it possible to specialize with char* pointer type parameter without typedef’ing it?

Answer 1

Sure. Try this:

template<>
void foo<char*>(char* const& value) {...}

This is because const char* means pointer to const char, not a const pointer to char.