Re: C23 thoughts and opinions

Em 5/24/2024 5:19 PM, Keith Thompson escreveu:

Thiago Adams <thiago.adams@gmail.com> writes:

On 24/05/2024 16:45, Keith Thompson wrote:

Thiago Adams <thiago.adams@gmail.com> writes:

On 23/05/2024 18:49, Keith Thompson wrote:

error: 'constexpr' pointer initializer is not null
5 |     constexpr char * s[] = {"a", "b"};
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Then we were asking why constexpr was used in that case.

Why not?

>
When I see a constexpr I ask if the compiler is able to compute
everything at compile time. If not immediately it is a bad usage in my
view.

I don't understand.  Do you object because it's not *immediately
obvious* that everthing can be computed at compile time?  If so, why
should it have to be?

>
My understanding is that constexpr is a tip for the compiler. Does not
ensure anything. Unless you use where constant expression is required.
So I don't like to see constexpr  where I know it is not a constant
expression.

 Your understanding is incorrect.  "constexpr" is not a mere hint.

I think I can explain I little better
Let´s consider we have a compile time array of integers and a loop.
https://godbolt.org/z/e8cM1KGWT
#include <stdio.h>
#include <stdlib.h>
int main() {
     constexpr int a[] = {1, 2, 3, 4, 5, 6, 7, 8};
     for (int i = 0 ; i < sizeof(a)/sizeof(a[0]); i++)
     {
         printf("%d", a[i]);
     }
}
What the programmer expected using a constant array in a loop?
The loop is in runtime, unless the compiler expanded the loop into 8 calls using constant expressions. But this is not the case.
This was the usage of constexpr I saw but with literal strings.
So, the array a is not used as constant even if it has constexpr.