Re: constexpr keyword is unnecessary

Em 10/13/2024 6:52 AM, Bart escreveu:

On 12/10/2024 22:43, Thiago Adams wrote:

Em 10/12/2024 10:53 AM, Bart escreveu:

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It depends on what constexpr means.

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It means the initialization expression must be evaluated at compilation.
Also the declarator can be used in another expression as constant.
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Sample
constexpr int a = 1;
constexpr int b = a+1;
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But the expression is always something the compiler need to check if is constant or not.
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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)
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For instance:
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const int a = 1;
const int b = a+1;
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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.
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 So it looks like const/constexpr do different things, for example:
     const int a = rand();              // OK
   constexpr int b = rand();          // error
 

Yes.
constexpr is like - "require the initializer to be a constant expression." But the compiler will have to check it anyway.
What I am suggesting is:
If the compiler is going to check anyway, after check, if the initializer is a constant expression, then transfer this property "compile time" to the variable automatically. Then no need for a new keyword.
One justification for the usage of constexpr could be , if constexpr is not present then we can skip this compile time check. But most C compiler will do optimization anyway.

How about this:
    static int x;
   const int* p = &x;                // OK
   constexpr int* q = &x;            // ??
 q's value isn't known at compile-time.
 

Addresses can be only initialized with 0 in C.
constexpr int* q = 0; //ok