Topic: Allow to access member typedefs of a type from a
Author: Walter Heisenbug <walter.heisenbug@outlook.com>
Date: Tue, 14 Mar 2017 18:03:02 +0000
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Currently to access a member type from a reference type means you have to d=
o all the dancing with removing the reference.=20
template<typename T>
void f(T&& x)
{
=C2=A0=C2=A0=C2=A0 typedef typename std::remove_reference_t<T>::member_type=
mem;
=C2=A0=C2=A0=C2=A0 mem y;
=C2=A0=C2=A0=C2=A0 use(std::forward<T>(x), y);
}
This could be simplified, if we allow to access the member types of a type =
from a reference types:
template<typename T>
void f(T&& x)
{
typedef typename T::member_type mem;
mem y;
use(std::forward<T>(x), y);
}
I'm not aware of any potential problems, since the reference types don't ha=
ve any member types, so there's no ambiguity with the existing code.=20
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Author: Jakob Riedle <jakob.riedle@gmail.com>
Date: Thu, 20 Apr 2017 00:57:25 -0700 (PDT)
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I quite like this idea and I had this problem myself. The reason why
compilers reject this code is because
'T is not a class, struct, or union type'. I don't know whether it is a
good idea to consider *reference to class *a *class type.*
It would be very misleading to do this if we allow references to class
types to be usable as if they were the referenced class:
struct A{};
struct B : A& {};
*References to objects* can be and are treated exactly like the referenced
object itself.
Note that this only concerns the *run-time*.
On the other hand a *reference type* is not a class, it is ... a reference.
IMHO it is very consequent, that e.g. T::r is not resolvable for T being a
reference type,
because its not a class, struct or union type.
Cheers
Jakob
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<div dir=3D"ltr">I quite like this idea and I had this problem myself. The =
reason why compilers reject this code is because<div>'<span style=3D"ba=
ckground-color: rgb(245, 245, 245); color: rgb(51, 51, 51); font-family: Me=
nlo, Monaco, Consolas, "Courier New", monospace; font-size: 13.33=
33px; white-space: pre;">T is not a class, struct, or union type'</span=
>. I don't know whether it is a good idea to consider <i>reference to c=
lass </i>a <i>class type.</i></div><div>It would be very misleading to do t=
his if we allow references to class types to be usable as if they were the =
referenced class:</div><div class=3D"prettyprint" style=3D"background-color=
: rgb(250, 250, 250); border-color: rgb(187, 187, 187); border-style: solid=
; border-width: 1px; word-wrap: break-word;"><code class=3D"prettyprint"><d=
iv class=3D"subprettyprint"><span style=3D"color: #008;" class=3D"styled-by=
-prettify">struct</span><span style=3D"color: #000;" class=3D"styled-by-pre=
ttify"> A</span><span style=3D"color: #660;" class=3D"styled-by-prettify">{=
};</span><span style=3D"color: #000;" class=3D"styled-by-prettify"><br></sp=
an><span style=3D"color: #008;" class=3D"styled-by-prettify">struct</span><=
span style=3D"color: #000;" class=3D"styled-by-prettify"> B </span><span st=
yle=3D"color: #660;" class=3D"styled-by-prettify">:</span><span style=3D"co=
lor: #000;" class=3D"styled-by-prettify"> A</span><span style=3D"color: #66=
0;" class=3D"styled-by-prettify">&</span><span style=3D"color: #000;" c=
lass=3D"styled-by-prettify"> </span><span style=3D"color: #660;" class=3D"s=
tyled-by-prettify">{};</span><span style=3D"color: #000;" class=3D"styled-b=
y-prettify"><br></span></div></code></div><div><br></div><div><b>References=
to objects</b> can be and are treated exactly like the referenced object i=
tself.</div><div>Note that this only concerns the <b>run-time</b>.</div><di=
v>On the other hand a <i>reference type</i> is not a class, it is ... a ref=
erence.</div><div><br></div><div>IMHO it is very consequent, that e.g. <spa=
n style=3D"background-color: rgb(245, 245, 245); color: rgb(51, 51, 51); fo=
nt-family: Menlo, Monaco, Consolas, "Courier New", monospace; fon=
t-size: 13.3333px; white-space: pre;">T::r</span>=C2=A0is not resolvable fo=
r=C2=A0<span style=3D"background-color: rgb(245, 245, 245); color: rgb(51, =
51, 51); font-family: Menlo, Monaco, Consolas, "Courier New", mon=
ospace; font-size: 13.3333px; white-space: pre;">T</span>=C2=A0being a refe=
rence type,</div><div>because its not a class, struct or union type.</div><=
div><br></div><div>Cheers</div><div>Jakob</div></div>
<p></p>
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Author: Dan Raviv <dan.raviv@gmail.com>
Date: Thu, 27 Apr 2017 13:23:22 -0700 (PDT)
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+1 for this non-breaking feature which has a similar precedent
(cv-qualifiers listed in Richard's answer), and which would simplify
writing (and *reading*) generic code.
Cheers,
Dan
On Tuesday, March 14, 2017 at 7:03:06 PM UTC+1, Walter Heisenbug wrote:
>
> Currently to access a member type from a reference type means you have to
> do all the dancing with removing the reference.
>
> template<typename T>
> void f(T&& x)
> {
> typedef typename std::remove_reference_t<T>::member_type mem;
> mem y;
> use(std::forward<T>(x), y);
> }
>
> This could be simplified, if we allow to access the member types of a type
> from a reference types:
>
> template<typename T>
> void f(T&& x)
> {
> typedef typename T::member_type mem;
> mem y;
> use(std::forward<T>(x), y);
> }
>
> I'm not aware of any potential problems, since the reference types don't
> have any member types, so there's no ambiguity with the existing code.
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<div dir=3D"ltr">+1 for this non-breaking feature which has a similar prece=
dent (cv-qualifiers listed in Richard's answer), and which would simpli=
fy writing (and *reading*) generic code.<br><br>Cheers,<div>Dan</div><div><=
br>On Tuesday, March 14, 2017 at 7:03:06 PM UTC+1, Walter Heisenbug wrote:<=
blockquote class=3D"gmail_quote" style=3D"margin: 0;margin-left: 0.8ex;bord=
er-left: 1px #ccc solid;padding-left: 1ex;">Currently to access a member ty=
pe from a reference type means you have to do all the dancing with removing=
the reference.=20
<br>
<br>template<typename T>
<br>void f(T&& x)
<br>{
<br>=C2=A0=C2=A0=C2=A0 typedef typename std::remove_reference_t<T>::<=
wbr>member_type mem;
<br>=C2=A0=C2=A0=C2=A0 mem y;
<br>=C2=A0=C2=A0=C2=A0 use(std::forward<T>(x), y);
<br>}
<br>
<br>This could be simplified, if we allow to access the member types of a t=
ype from a reference types:
<br>
<br>template<typename T>
<br>void f(T&& x)
<br>{
<br>=C2=A0 =C2=A0 typedef typename T::member_type mem;
<br>=C2=A0 =C2=A0 mem y;
<br>=C2=A0 =C2=A0 use(std::forward<T>(x), y);
<br>}
<br>
<br>I'm not aware of any potential problems, since the reference types =
don't have any member types, so there's no ambiguity with the exist=
ing code. </blockquote></div></div>
<p></p>
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