Trying to pass a constexpr lambda and use it to explicitly specify returning type

Question

I would like to use a function and pass a constexpr lambda. However, it only compiles successfully if I let the type be deduced through auto. Explicitl

Answer 1

Parameters to constexpr functions are not themselves constexpr objects - so you cannot use them in constant expressions. Both of your examples returning arrays are ill-formed because there is no valid call to them.

To understand why, consider this nonsense example:

struct Z { int i; constexpr int operator()() const { return i; }; };

template <int V> struct X { };
template <typename F> constexpr auto foo(F f) -> X<f()> { return {}; }

constexpr auto a = foo(Z{2});
constexpr auto b = foo(Z{3});

Z has a constexpr call operator, and this is well-formed:

constexpr auto c = Z{3}();
static_assert(c == 3);

But if the earlier usage were allowed, we'd have two calls to foo<Z> that would have to return different types. This could only fly if the actual value f were the template parameter.

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Note that clang compiling the declaration is not, in of itself, a compiler error. This is a class of situations that are ill-formed, no diagnostic required.