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From: Bart <bc@freeuk.com>
Newsgroups: comp.lang.c
Subject: Re: constexpr keyword is unnecessary
Date: Sun, 13 Oct 2024 10:52:55 +0100
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On 12/10/2024 22:43, Thiago Adams wrote:
> Em 10/12/2024 10:53 AM, Bart escreveu:
>>
>> It depends on what constexpr means.
> 
> It means the initialization expression must be evaluated at compilation.
> Also the declarator can be used in another expression as constant.
> 
> Sample
> constexpr int a = 1;
> constexpr int b = a+1;
> 
> But the expression is always something the compiler need to check if is 
> constant or not.
> 
> So, what I suggest is if the init expression is a constant expression 
> (something we know at compile time) then the declarator is real 
> constexpr (no need for a new keyword)
> 
> For instance:
> 
> const int a = 1;
> const int b = a+1;
> 
> 
> Even if not constant, the compiler can do at compile time.
> for instance.
> int a = 1+2;
> So this has to be done anyway.
> The only difference is that using 'a' cannot be used as constant in 
> another expression.
> 
> 

So it looks like const/constexpr do different things, for example:


   const int a = rand();              // OK
   constexpr int b = rand();          // error

How about this:

   static int x;
   const int* p = &x;                // OK
   constexpr int* q = &x;            // ??

q's value isn't known at compile-time.