Template casting operator

[Jonathan Turkanis]

Re: Template casting operator


"Jonathan Turkanis" <technews@kanga roologic.com> wrote in message
news:bvua5m$t9f 91$1@ID-216073.news.uni-berlin.de...[color=blue]
>
> "Makhno" <root@127.0.0.1 > wrote in message
> news:bvu9lt$lk$ 1@newsg4.svr.po l.co.uk...[color=green][color=darkred]
> > > Did you try using converting constructors instead?[/color]
> >
> > Yes, worked straight away. You're right - perhaps I don't need the[/color]
> cast[color=green]
> > operator at all.
> >[color=darkred]
> > > You probably need a templated assignment operator too, if you do[/color][/color]
> this.[color=green]
> >
> > No, as
> > Vector<float> pf=pd;
> > and
> > Vector<float> pf(pd);
> > are equivalent;
> >[/color][/color]
[color=blue]
>
> I didn't suggest a templated assignement operator first because your
> example didn't require it. But if you want a general-purpose
> substitute for the conversion operator, you need it.
>[/color]

I spoke too soon. It's non needed here either. But it's not because
of copy initialization.

Jonathan

[Makhno]

Re: Template casting operator

> But that's just one use of the implicit conversion.[color=blue]
>
> How about:
>
> Vector<double> pd;
> Vector<float> pf;
> // More stuff here.
> pf = pd;
>
> ?[/color]

Hmm, this still seems to work, how peculiar. Vector<> doesn't have an
assignment operator.

[Makhno]

Re: Template casting operator

> I spoke too soon. It's non needed here either. But it's not because[color=blue]
> of copy initialization.[/color]

why then?

[Victor Bazarov]

Re: Template casting operator

"Jonathan Turkanis" <technews@kanga roologic.com> wrote...[color=blue]
>
> "Makhno" <root@127.0.0.1 > wrote in message
> news:bvu9lt$lk$ 1@newsg4.svr.po l.co.uk...[color=green][color=darkred]
> > > Did you try using converting constructors instead?[/color]
> >
> > Yes, worked straight away. You're right - perhaps I don't need the[/color]
> cast[color=green]
> > operator at all.
> >[color=darkred]
> > > You probably need a templated assignment operator too, if you do[/color][/color]
> this.[color=green]
> >
> > No, as
> > Vector<float> pf=pd;
> > and
> > Vector<float> pf(pd);
> > are equivalent;
> >[/color]
>
> But that's just one use of the implicit conversion.
>
> How about:
>
> Vector<double> pd;
> Vector<float> pf;
> // More stuff here.
> pf = pd;
>
> ?
>
> I didn't suggest a templated assignement operator first because your
> example didn't require it. But if you want a general-purpose
> substitute for the conversion operator, you need it.[/color]

Why? 'pf = pd' expression will result into the use of compiler-
generated assignment operator and a construction of a temporary
on the right side, no? It should be equivalent to

{ Vector<float> temp(pd); pf = temp; }

No special assignment operator is needed here.

Correct me if I am wrong.

V

[Jonathan Turkanis]

Re: Template casting operator


"Victor Bazarov" <v.Abazarov@com Acast.net> wrote in message
news:37yUb.1814 15$Rc4.1339653@ attbi_s54...[color=blue]
> "Jonathan Turkanis" <technews@kanga roologic.com> wrote...[color=green]
> >[/color][/color]
[color=blue]
> Why? 'pf = pd' expression will result into the use of compiler-
> generated assignment operator and a construction of a temporary
> on the right side, no? It should be equivalent to
>
> { Vector<float> temp(pd); pf = temp; }
>
> No special assignment operator is needed here.
>
> Correct me if I am wrong.
>[/color]

You're not -- I was :(.

Jonathan

[Jonathan Turkanis]

Re: Template casting operator


"Makhno" <root@127.0.0.1 > wrote in message
news:bvuak9$253 $1@newsg1.svr.p ol.co.uk...[color=blue][color=green]
> > I spoke too soon. It's non needed here either. But it's not[/color][/color]
because[color=blue][color=green]
> > of copy initialization.[/color]
>
> why then?
>[/color]

See Victor's post.

Jonathan

[Victor Bazarov]

Re: Template casting operator

"Makhno" <root@127.0.0.1 > wrote...[color=blue][color=green]
> > But that's just one use of the implicit conversion.
> >
> > How about:
> >
> > Vector<double> pd;
> > Vector<float> pf;
> > // More stuff here.
> > pf = pd;
> >
> > ?[/color]
>
> Hmm, this still seems to work, how peculiar. Vector<> doesn't have an
> assignment operator.[/color]

Yes, it does. If you don't declare one yourself, the compiler will
do it for you.