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Macros are often used for text substitution. In my case, I need to conditionally wipe out some keywords, so that compilation will be possible in compilers that don't have the specific feature.

Specifically I've been looking into cpp11 range where this snippet comes from 

template <typename C>
struct has_size {
    template <typename T>
    static constexpr auto check(T*) -> // problem in VS2013
        typename std::is_integral<
            decltype(std::declval<T const>().size())>::type;
// .. some more stuff
};

I'm providing this as the example that spawned the question. In the above code I ended up doing

template <typename T>
static 
#if COMPILE_WITH_GCC_NEW_ENOUGH
    constexpr 
#endif
auto check(T*) ->

because there were other parts where constexpr needed to be replaced with const in order to compile.

What I'm asking is a way to say eg

#define Constexpr ?????????

so that it would be replaced by constexpr in gcc compilations and textually nothing in VS compilations.

Nikos Athanasiou
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1 Answers1

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Certainly, simply:

#define constexpr

will expand constexpr to nothing. You can wrap this macro definition inside an appropriate #if for your compiler that doesn't support this.

Sometimes people make the "empty" expansion more explicit:

#define constexpr /* nothing */
Greg Hewgill
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  • I have seen this so many times in include guards that I somehow deduced it only _defines_ a symbol for checking with `#ifndef` or `#ifdef`. Thanks for the quick response – Nikos Athanasiou Apr 28 '14 at 22:01