Question

Say I have a macro like this:

#define SET_TYPE_NAME(TYPE, NAME) \
    template<typename T>          \
    std::string name();           \
                                  \
    template<>                    \
    std::string name<TYPE>() {    \
        return NAME;              \
    }

This won’t work if I pass it a template that has more than one parameter, because the comma in the <int, int> is interpreted as separating the macro arguments, not the template arguments.

SET_TYPE_NAME(std::map<int, int>, "TheMap")
// Error: macro expects two arguments, three given

This problem seems to be solved by doing this:

SET_TYPE_NAME((std::map<int, int>), "TheMap")

But now another problem arises, one that I really did not expect:

 template<>
 std::string name<(std::map<int, int>)>()
 // template argument 1 is invalid

It seems that the extra parentheses make the template argument invalid. Is there any way around this?

Answer 1

Besides typedef, you could switch the order of the arguments and use variadic macros (requires C99 or C++11-compatible compiler):

#define SET_TYPE_NAME(NAME, ...) \
template<typename T>          \
std::string name();           \
                              \
template<>                    \
std::string name<__VA_ARGS__>() {    \
    return NAME;              \
}

…

SET_TYPE_NAME("TheMap", std::map<int, int>)