Why is `std::is_constructible_v == false`

Question

Reproducible example:

#include <type_traits>
static_assert(std::is_constructible_v<int[2], int, int>, "fails against my expectations");

I tested this with clang 5 and gcc 7.

Why do you expect it to be true? An array does not have a constructor. From [this](http://en.cppreference.com/w/cpp/types/is_constructible): "*If `T` is an object or reference type and the variable definition `T obj(std::declval()...);` is well-formed, provides the member constant value equal to `true`. **In all other cases, value is `false`**.*" `int obj[2](int, int)` is not well-defined. Dont confuse construction with aggregate initialization. — Remy Lebeau, Nov 28 '17 at 07:50

gsamaras · Accepted Answer · 2017-11-28T08:14:15.530

8

From the ref:

If T is an object or reference type and the variable definition T obj(std::declval<Args>()...); is well-formed, provides the member constant value equal to true. In all other cases, value is false.

In your example T obj(std::declval<Args>()...); is not well-formed.

This is because int[2] is a plain array, which doesn't have any constructor whatsoever.

As a result, this:

int obj[2](int, int);

is ill-formed.

Arrays are aggregates, thus aggregate initialization comes into play here, not construction.

edited Nov 28 '17 at 08:14

answered Nov 28 '17 at 07:50

gsamaras

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Do you know if there is a way to check if a type is constructible with aggregates? Ie, is_aggregate_initializable would be true? – BWG Jul 18 '22 at 04:36

score 2 · Answer 2 · answered Nov 28 '17 at 07:49

The purpose and definition of std::is_constructible is to check if an object of the specified type may be constructed as in:

T obj(std::declval<Args>()...);

And the above is simply not well-formed for an array. An array doesn't have any constructors, it is an aggregate and should be initialized with aggregate initialization.

Why is `std::is_constructible_v == false`

2 Answers2