Find out the type of __VA_ARGS__ in a variadic macro

Question

Supposedly, I have a variadic macro (e.g., MY_MACRO(...)) and I call it the following way:

MY_MACRO(std::pair<int, int> const &p)

Now, __VA_ARGS__ in my macro's body would be std::pair<int, int> const &p.

Is there a way to figure out the type of __VA_ARGS__?

Presumably, I would be grateful if something like decltype(std::pair<int, int> const &p) worked and yield std::pair<int, int> const&, so in my variadic macro's body decltype(__VA_ARGS__) would yield std::pair<int, int> const& as well. Unfortunately this doesn't work.

Preprocessor-macros is C++ stuff, like everything else inherited and / or adopted from C. — Deduplicator, Oct 03 '18 at 22:29
Macro arguments are arbitrary strings, which needn't be parsable. So speaking of the type of the argument makes little sense. — , Oct 03 '18 at 22:34
@YvesDaoust I know, but I was hoping for some evil sourcery... — 101010, Oct 03 '18 at 22:34
Does the [comma really work in your case?](https://stackoverflow.com/q/24793706/1708801) — Shafik Yaghmour, Oct 03 '18 at 22:53
@ShafikYaghmour: It's a list of two arguments, but that can work with `__VA_ARGS__` — MSalters, Oct 03 '18 at 23:10

score 5 · Accepted Answer · answered Oct 03 '18 at 22:34

5

You can use __VA_ARGS__ as a lambda parameter, then convert that lambda to a function pointer and extract the parameter type:

template <typename T> struct func_param {};
template <typename T> struct func_param<void(*)(T)> {using type = T;};

#define FOO(...) \
    do \
    { \
        auto lambda = +[]([[maybe_unused]] __VA_ARGS__) {}; \
        using type = func_param<decltype(lambda)>::type; \
        /* Do something with `type`. */\
    } \
    while (0);

answered Oct 03 '18 at 22:34

HolyBlackCat

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Note using [the fact that lambda w/o capture can be converted to function pointer](https://stackoverflow.com/q/28746744/1708801) – Shafik Yaghmour Oct 03 '18 at 22:55
Yes, and also the trick that applying unary `+` to such a lambda forces the conversion, since there is no overloaded `operator+` matching the lambda type, but the pointer to function is a valid operand of the built-in meaning of unary `+`. – aschepler Oct 03 '18 at 23:08

Find out the type of __VA_ARGS__ in a variadic macro

1 Answers1