brace initialization of subclasse of array

[Ralf Goertz]

Hi,

why does brace initialization of a subclass of array fail and how do I
fix this?


#include <array>

//typedef std::array<double,3> Space; //with this line it works
struct Space : public std::array<double,3>{};

int main() {
Space x={{1,2,3}}; //fails when Space is subclass of array
return 0;
}

[Alf P. Steinbach]

It's no longer an aggregate (simple class with no bases or user-provided
constructors). List-initialization reduces to aggregate initialization
when the class is an aggregate. Otherwise various special cases are
considered, e.g. initializing a `std::initializer_list`, and in your
case constructors are considered, and none were found.

You can simply provide a constructor that takes a `std::initializer_list`.

`std::array` stems from Boost, from the time before
`std::initializer_list`, and was designed for old C++ and C aggregate
initialization.


Cheers & hth.,

- Alf

[Ralf Goertz]

Am Thu, 14 Sep 2017 10:48:19 +0200
schrieb "Alf P. Steinbach" <alf.p.stein...@gmail.com>:

> On 9/14/2017 10:18 AM, Ralf Goertz wrote:
> > Hi,
> >
> > why does brace initialization of a subclass of array fail and how
> > do I fix this?
> >
> >
> > #include <array>
> >
> > //typedef std::array<double,3> Space; //with this line it works
> > struct Space : public std::array<double,3>{};
> >
> > int main() {
> > Space x={{1,2,3}}; //fails when Space is subclass of array
> > return 0;
> > }
> >
>
> It's no longer an aggregate (simple class with no bases or
> user-provided constructors). List-initialization reduces to aggregate
> initialization when the class is an aggregate.

Hm, you mean to say it works with array because it doesn't have a
user-provided constructor so it reduces to aggregate initialization?
Which then fails in a subclass because it is no longer aggregate? But
the above program also fails to compile when I substitute array with
vector (but still works with the typedef). And vector does have
non-default constructors hence it is not aggregate. So why doesn't a
subclass of vector inherit its list-initialization constructors?

[Alf P. Steinbach]

If you want constructor inheritance then you have to specify that
explicitly, e.g. (off the cuff)

struct Foo
: std::vector<int>
{
using std::vector<int>::vector<int>; // Inherit constrcts
};

[Ralf Goertz]

Am Thu, 14 Sep 2017 16:14:31 +0200

schrieb "Alf P. Steinbach" <alf.p.stein...@gmail.com>:

> If you want constructor inheritance then you have to specify that
> explicitly, e.g. (off the cuff)

That cuff needs a link ;-)

> struct Foo
> : std::vector<int>
> {
> using std::vector<int>::vector<int>; // Inherit constrcts

error: a template-id may not appear in a using-declaration

> };

I don't seem to be able to fix that though. :-(

[Alf P. Steinbach]

Oh sorry.

#include <vector>

struct Foo
: std::vector<int>
{

using std::vector<int>::vector;
};

auto main()
-> int
{
Foo const v{ 1, 2, 3 };
}

There.

Another way is to define a name for the base class.

[asetof...@gmail.com]

Using
std::ftrytr:: Etc
As the line you wrote
it is for me horrible
There would be some other way to access in namespace

[Öö Tiib]

On Thursday, 14 September 2017 13:22:05 UTC+3, Ralf Goertz wrote:
> So why doesn't a
> subclass of vector inherit its list-initialization constructors?

Because language rules (IMHO fortunately) do not suggest anywhere that
subclass should somehow automatically inherit all constructors.