Constexpr lambda argument

Question

Is it possible to have a lambda with a constexpr argument? And is it possible to make the following example work?

ForEach function provided below calls a given lambda 3 times with index 0, 1, 2:

template <class Func, std::size_t... index>
inline constexpr void ForEach(Func && f, std::index_sequence<index...>)
{
    (f(index), ...);
}

template <class Func>
inline constexpr void ForEach(Func && f)
{
    ForEach(f, std::make_index_sequence<3>());
}

so the following code

ForEach([](size_t index)
{
    std::cout << index << ' ' << std::endl;
});

outputs 0, 1, 2.

But the following code that tries to print tuple elements requires index to be a constexpr:

auto t = std::make_tuple(1, 2.0, std::string("abc"));

ForEach([&t](size_t index)
{
    std::cout << std::get<index>(t) << ' ' << std::endl;
});

and thus does not compile, see live example. Is it possible to make index constexpr somehow?

EDIT1: There is a working example where a lambda argument is used as a template argument:

void Set(Tuple& val, size_t index, Variant const& elem_v)
{
    mp_with_index<std::tuple_size_v<Tuple>>(
        index,
        [&](auto I){
            std::visit([&](auto const& alt){
                if constexpr (std::is_assignable_v<
                        std::tuple_element_t<Tuple, I>,
                        decltype(alt)>)
                {
                    std::get<I>(val) = alt;
                } else {
                    throw /* something */;
                }
            }, elem_v);
        });
}

why does this compile, but my sample code does not?

No. Function arguments can't be treated as constexpr (not sure whether C++20 changes that). — Fureeish
– Fureeish, Commented Jul 8, 2019 at 15:16
@AlexeyStarinsky I is a std::integral_constant<std::size_t, (something)> which has a constexpr operator std::size_t() that returns that (something) — Artyer
– Artyer, Commented Jul 8, 2019 at 15:22

Barry · Accepted Answer · 2019-07-08 15:25:32Z

In this:

ForEach([&t](size_t index)
{
    std::cout << std::get<index>(t) << ' ' << std::endl;
});

index is not a constant expression. It's just a variable. Function parameters are not constexpr.

But if we tweaked ForEach somewhat (to work the same way as the example of mine that you linked):

template <class Func, std::size_t... index>
inline constexpr void ForEach(Func && f, std::index_sequence<index...>)
{
    (f(std::integral_constant<std::size_t, index>()), ...);
    // ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    //    instead of just index
}

ForEach([&t](auto index)
{
    std::cout << std::get<index>(t) << ' ' << std::endl;
});

Then this works because index is no longer size_t but rather different instances of std::integral_constant<size_t, V> for various V. That type looks something like:

template<class T, T v>
struct integral_constant {
    static constexpr T value = v;
    typedef T value_type;
    typedef integral_constant type; // using injected-class-name
    constexpr operator value_type() const noexcept { return value; }
    constexpr value_type operator()() const noexcept { return value; } //since c++14
};

Converting a std::integral_constant<size_t, V> to a size_t invokes the constepxr operator size_t(), which doesn't involve reading any state from this object itself (which is an empty type), hence it's allowed as a constant expression.

A different way of looking at it is that we're encoding the value in the type (which can be retrieved as a constant expression) rather than in the value (which cannot).

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