[Solved] Is there a way to define Variadic template macro?

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__VA_ARGS__ can be used multiple times, so you could write:

#define PARSE_FUNCTION(functionName, ...) \
    std::function<int(__VA_ARGS__)> m_##functionName() { \
        return std::function<int(__VA_ARGS__)>(functionName); \
    }

What is happening is just simple text substitution, whether the arguments is for a template or not won’t be checked by the preprocessor.

Actually any function objects can be implicitly converted to a std::function, so the cast can be omitted. Furthermore, if functionName refers to function pointers, the exact type can be easily inferred, that you don’t need the variadic macro at all:

#define PARSE_FUNCTION(functionName) \
    auto m_##functionName() \
        -> std::function<std::remove_pointer<decltype(functionName)>::type> \
    { \
        return functionName; \
    }

solved Is there a way to define Variadic template macro?