Re: constexpr keyword is unnecessary

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.