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?Explevoid UsePointer(int*);void foo(){ int var = init(); .../// do something with var  const var;  ++var; // should fail right?  UsePointer(&var);// cast away const???}Sent from my Galaxy
-------- Original message --------From: Bo Persson via Std-Proposals <std-proposals_at_[hidden]> Date: 2/16/25 12:54 PM (GMT+01:00) To: std-proposals_at_[hidden] Cc: Bo Persson <bo_at_[hidden]> Subject: Re: [std-proposals] Delayed const declaration On sön 2025-02-16 at 12:08, Tiago Freire via Std-Proposals wrote:> 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).-- Std-Proposals mailing listStd-Proposals_at_[hidden]://lists.isocpp.org/mailman/listinfo.cgi/std-proposals