Re: constexpr keyword is unnecessary

Em 10/13/2024 10:29 AM, Michael S escreveu:

On Sun, 13 Oct 2024 10:10:19 -0300
Thiago Adams <thiago.adams@gmail.com> wrote:
 

Em 10/13/2024 10:06 AM, Michael S escreveu:

On Sun, 13 Oct 2024 09:38:04 -0300
Thiago Adams <thiago.adams@gmail.com> wrote:
  

Em 10/13/2024 8:49 AM, Bonita Montero escreveu:

Am 13.10.2024 um 13:37 schrieb Thiago Adams:
     

Yes.
constexpr is like - "require the initializer to be a constant
expression." But the compiler will have to check it anyway.

>
I cannot understand why you are so militantly against this
new language feature that can be understood in 10 seconds.
     

>
I have seen code like this:
>
void func()
{
      constexpr int c = 1;
      f(c);
}
>
For some reason, people believe that adding constexpr will
magically improve optimization. In reality, it doesn't change
anything compared to const and often reflects a misunderstanding
of how the compiler works. As a result, I end up having to explain
it. In this sense, constexpr is viral and spreads confusion.
 

>
I see constexpr primarily as a way to enable use of functions from
math.h in static initializers.
  

>
Maybe you are thinking in C++? C does not have compile time functions.
>

 I'd expect that in the next standard a wide subset of math functions
would be allowed in constexp.
In C++ they become constexpr in C++23. If no unexpected difficulties
shows up in C++ then C would be next.
 

Let´s say we have
double  i = sin(1);
adding constexpr
constexpr double i = sin(1);
makes sin(1) to be evaluated in compile time.
**BUT**, I may want to compute sin(1) even without making i const.
double i = sin(1);
This shows another incoherence.
The desire of pre compute the function in compile time should not be dependent of the declarator being const.
So, in this case I would create a _Static_eval.
double i = _Static_eval(sin(1));
This feature is not competing with const creating two ways of doing something. (like constexpr is doing)
This is a complement to the language to force compile time computation in cases the expression is not required to be a constant expression.