std::apply and constant expression?

Question

I tried code below in Wandbox:

#include <array>
#include <iostream>
#include <tuple>
#include <typeinfo>
#include <functional>
#include <utility>


int main()
{
    constexpr std::array<const char, 10> str{"123456789"};
    constexpr auto foo = std::apply([](auto... args) constexpr { std::integer_sequence<char, args...>{}; } , str);
    std::cout << typeid(foo).name();
}

and the compiler told me that args... are not constant expression. What's wrong?

Blowfish

Unless this has changed in C++1z, you cannot have constexpr function parameters. That is, every function must assume it may be called with runtime parameters, and then your lambda doesn't make sense. – Inception

@Inception What a pity… – Blowfish

Answer 1

All constexpr functions must be valid both constexpr and not, even if marked constexpr.

There is a proposal for a constexpr literal that passed characters as non type template parameters. Then "hello"_bob could expand directly to a parameter pack.

Another approach is you can pass std::integral_constant<T, t> to the lambda through some mechanism, like my indexer. Then converting to T is constexpr even tho the variable is not. This does not help you with "hello" to a sequence.

Fussy

Answer 2

Function parameters cannot be labeled constexpr. As such, you cannot use them in places that require constant expressions, like non-type template arguments.

To do the kind of thing you're trying to do would require some kind of compile-time string processing, based around template arguments.

Wappes

Answer 3

What you want can be done without std::apply:

#include <array>
#include <iostream>
#include <tuple>
#include <typeinfo>
#include <functional>
#include <utility>
#include <type_traits>

template <std::size_t N, class = std::make_index_sequence<N>>
struct iterate;

template <std::size_t N, std::size_t... Is>
struct iterate<N, std::index_sequence<Is...>> {
    template <class Lambda>
    constexpr auto operator()(Lambda lambda) {
        return lambda(std::integral_constant<std::size_t, Is>{}...);
    }
};

int main()
{
    constexpr std::array<const char, 10> str{"123456789"};
    constexpr auto foo = iterate<str.size()>{}([](auto... is) constexpr { return std::integer_sequence<char, str[is]...>{}; });
    std::cout << typeid(foo).name();
}

[live demo]

Chessboard

I actually now think it shouldn't but not sure yet... :/ – Chessboard

I'm more concerned about using arguments in a constant expression. As per the other answers I'd have expected it to ensure this is also callable with runtime arguments. – Inception

Though I don't have such doubts with this code. It's a clever idea! – Inception

I think this one is actually ok. std::integral_constant has its overload of constexpr operator T() allowing it to be used in constexpr context. Or am I mising something? – Chessboard

The question if str declared in main can be used in context of lambda is on the other hand still opened.. – Chessboard

You seem to be right about integral constants. eel.is/c++draft/expr.prim.lambda#10 and eel.is/c++draft/expr.prim.lambda#13 suggest that your code is perfectly okay – Inception

Nice founding! Thank you! – Chessboard

You can even make "factory" for it, but currently gcc crashed on this melpon.org/wandbox/permlink/JRvLX98vakSP1k2k . – Bondstone

@Orient Nice test for compiler you've made... Have you watched what syntax element caused compiler crash was it constexpr lambda or auto template parameter...? – Chessboard

Namely auto const & template parameter. If it slightly modified to be template< std::size_t size, char const (& value)[size] >, then all is OK, but then I need to provide size parameter manually. – Bondstone

@Orient one word - I've just noticed you used gcc clang extension on string literal, while I'm opened on that, some people would probably like to wait for when it become official c++ feature... – Chessboard

@Chessboard I submitted the bug, but it's status is not changed yet to resolved. That extension is quite old. Don't know is there a proposal for it. – Bondstone

@Chessboard Which wandbox compiled fine? This still can't. – Bondstone

Let us continue this discussion in chat. – Chessboard