c++ - Fold expressions with arbitrary callable?

Question

Looking over the C++17 paper on folds, (and on cppreference), I'm confused as to why the choice was made to only work with operators? At first glance it seems like it would make it easier to expand (... + args) by just shoving a + token between the elements of args, but I'm unconvinced this is a great decision.

Why can't a binary lambda expression work just as well and follow the same expansion as the latter above? It's jarring to me that a fold syntax would be added to a language without support for arbitrary callables, so does the syntax allow a way to use them that I'm just not seeing?

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Update: This works for a variadic min() function with clang

template <typename T>
struct MinWrapper {
    const T& obj;
};

template <typename T, typename U, typename V=std::common_type_t<T,U>>
constexpr MinWrapper<V> operator%(
        const MinWrapper<T>& lhs, const MinWrapper<U>& rhs) {
    return {lhs.obj < rhs.obj ? lhs.obj : rhs.obj};
}


template <typename... Ts>
constexpr auto min(Ts&&... args) {
    return (MinWrapper<Ts>{args} % ...).obj;
}

question from:https://stackoverflow.com/questions/27582862/fold-expressions-with-arbitrary-callable

Answer 1

First of all I'm happy that what I wrote works in clang (I see your update implements 3 out of the 4 steps I mention). I have to give credit to Nick Athanasiou for this technique that discussed this with me way before writing this.

The reason I mention this now is because I was informed he released a library (in the boost library incubator) that implements this stuff; you can find related documentation here. It seems the initial idea (that we both use here) and allowed code like this:

(Op<Max>(args) + ...); // Op is a function producing the custom fold type

was left out in favor of lazy evaluation and stateful operators (or not included yet, can't know for sure).