array member initialization of user defined types

Question

g++ 4.7 supports array member initialization and I started playing with it.

The code below does not compile.

struct A
{
   A(int){};
   A(const A&) = delete;
   A& operator=(const A&) = delete;
   ~A(){};
};

struct B
{
   B():
      a{{0},{1}}
   {};
   A a[2];
};

B b;

The error message with gcc 4.8 (prerelease) is:

n.cc: In constructor ‘B::B()’:
n.cc:12:20: error: use of deleted function ‘A::A(const A&)’
           a{{0},{1}}
             ^
n.cc:4:8: error: declared here
        A(const A&) = delete;
        ^

Is there a way to make this code work? I can't easily change the contructors,destructor of A. I seem to need a move-constructor or copy-constructor to initialize the array, but this seems counter-intuitive, since all I really want is in-place construction.

It works if I split a[2] in 2 members a0 and a1, and construct them separately. This looks fishy however.

':' in a constructor uses copy constructors to assign values, and you delete it in A thus giving you an error. — Syntactic Fructose
– Syntactic Fructose, Commented Dec 14, 2012 at 21:06

Community · Accepted Answer · 2017-05-23 11:52:41Z

4

Within list-initialization (8.5.4p1) of an aggregate (8.5.1), the form of initialization performed on the elements of the aggregate is copy-initialization (8.5.1p2), even if the initialization is direct-list-initialization:

When an aggregate is initialized by an initializer list, as speciﬁed in 8.5.4, the elements of the initializer list are taken as initializers for the members of the aggregate, in increasing subscript or member order. Each member is copy-initialized from the corresponding initializer-clause.

However, just because the form of initialization performed is copy-initialization doesn't mean that a copy occurs. As per copy-list-initialization of non-copyable types, copy-list-initialization should be identical to direct-list-initialization with the exception that explicit constructors are not allowed.

edited May 23, 2017 at 11:52

CommunityBot

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answered Dec 14, 2012 at 21:21

ecatmur

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Comments

ildjarn · Accepted Answer · 2012-12-14 21:27:43Z

2

Arrays are aggregates, and aggregate initialization always uses copy initialization. C++11 §8.5.1/1:

An aggregate is an array or a class with no user-provided constructors, no brace-or-equal-initializers for non-static data members, no private or protected non-static data members, no base classes, and no virtual functions.

§8.5.1/2:

When an aggregate is initialized by an initializer list, as specified in 8.5.4, the elements of the initializer list are taken as initializers for the members of the aggregate, in increasing subscript or member order. Each member is copy-initialized from the corresponding initializer-clause. …

(Emphasis mine.)

Additionally, the compiler does not implicitly generate a move constructor if a user-declared copy constructor is present (§12.8/9); because you have a user-declared copy constructor that is defined as deleted, A has neither a copy nor a move constructor. Explicitly adding a move constructor works:

struct A
{
   A(int) { }
   A(A const&) = delete;
   A& operator = (A const&) = delete;
   A(A&&) = default;
   ~A() = default;
};

struct B
{
   B() : a{{0}, {1}} { }
   A a[2];
};

int main()
{
   B b;
}

_{Online demo}

This is about as close to what you want as you're going to get.

edited Dec 14, 2012 at 21:27

answered Dec 14, 2012 at 21:12

ildjarn

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1 Comment

mirk Over a year ago

Not the answer I was hoping for, but thanks for the detailed explanation.

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