Tiago,
When you delay declare your variable as const, for the rest of the code, how can it know that it should use the const version of the non const?
Exple
void UsePointer(int*);
void foo()
{
int var = init();
.../// do something with var
const var;
++var; // should fail right?
UsePointer(&var);// cast away const???
}
-------- Original message --------
From: Bo Persson via Std-Proposals <std-proposals@lists.isocpp.org>
Date: 2/16/25 12:54 PM (GMT+01:00)
To: std-proposals@lists.isocpp.org
Cc: Bo Persson <bo@bo-persson.se>
Subject: Re: [std-proposals] Delayed const declaration
> If you are going to do that, perhaps we can simplify the syntax a little
> bit.
>
> Instead of having to declare [mutable const] which makes the variable
> open for writes like any other regular variable.
>
> We could just omit [mutable const] and deduce that after the fact
> depending on either or not a close statement is used, instead of:
>
> [mutable const] int x = 0;
>
> const x;
>
> you would simplify it as:
>
> int x = 0;
>
> const x;
>
The two-step declaration has the disadvantage that it invites typos. Or
when you change the name of x, you might forget the const declaration
half a page further down. Now some other x is const?
This reminds me of a similar problems with goto, where added code before
and after the label affects what the goto means (in a totally different
location).
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