Why do I need to member-initialize a non-static array member in a constexpr class?

Question

The following code desperately needs : values() to compile, at least in ideone::C++14:

#include <iostream>

template<int N>
struct Table
{
    constexpr Table() : values()
    {
        for (auto i = 0; i < N; ++i)
        {
            values[i] = i * i * i;
        }
    }
    int values[N];
};

int main() {
    constexpr auto a = Table<1000>();
    for (auto x : a.values)
        std::cout << x << '\n';
}

But why? I had thoughts along "values could also be initialized in a non-constexpr way and values() does explicitly say that we initialize it in a constexpr-compliant manner". But is not omitting : values() just as clear?

clang gives a strange message: assignment to object outside its lifetime is not allowed in a constant expression for the = in the constructor of Table if I omit values(). — Holt
– Holt, Commented May 24, 2016 at 12:45
"for the constructor of a class or struct, every base class sub-object and every non-variant non-static data member must be initialized. " — krOoze
– krOoze, Commented May 24, 2016 at 13:36
But it is initialized anyways, just Not explicit in any case. It works for non-constexpr classes — IceFire
– IceFire, Commented May 24, 2016 at 13:43

StoryTeller - Unslander Monica · Accepted Answer · 2016-05-24 14:10:12Z

5

Consider the semantics.

Omitting the member from the initialization list will perform default initialization, which in this case leaves the array with unspecified values. That negates the purpose of a constexpr.

Value initializing the array performs zero initialization on each array element (since this is an array of built in types).

answered May 24, 2016 at 14:10

StoryTeller - Unslander Monica

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3 Comments

IceFire Over a year ago

this indeed makes sense and I did not know that

Johannes Schaub - litb Over a year ago

Still, why is it not allowed to initialize the elements in the for loop? Afterwards, all array elements have values

StoryTeller - Unslander Monica Over a year ago

@JohannesSchaub-litb, For simplicity of implementation, I think. Judging by the quote Serge put bellow, the committee may have thought the following the c'tor chain may be less intractable than tracking an arbitrary c'tor body.

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

Simply because it is required by standard. Draft n4296 for current C++ standard states at :

7.1.5 The constexpr specifier [dcl.constexpr] §4 (emphasize mine):

4 The definition of a constexpr constructor shall satisfy the following constraints:
...

In addition, either its function-body shall be = delete, or it shall satisfy the following constraints:

(4.4) — either its function-body shall be = default, or the compound-statement of its function-body shall satisfy the constraints for a function-body of a constexpr function;
(4.5) — every non-variant non-static data member and base class sub-object shall be initialized (12.6.2);
...

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 24, 2016 at 14:01

Serge Ballesta

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1 Comment

StoryTeller - Unslander Monica Over a year ago

Perhaps the standard should clarify really that default initialization does not apply as "initialized". However, having an unspecified value really should preclude from being a constexpr, so I guess it follows logically.

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