CODEWITHSUNDEEP
c++constant-expression
user2953119
user2953119

Reputation:

Constant initialization in a dynamic way

I think that the the variable declared as const applies only Static Initialization. I've written the following:

#include <cstdlib> 
#include <iostream>

struct A{ };

const A *i = new A();

int main(){ }

and it works fine.

Ideone

But I expected that the the code is invalid because new A() is a new-expression and it is not a constant expression. Actually: sec. 5.19/2 N3797:

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:
[...]
— a new-expression (5.3.4);

[...]

and

A constant expression is either a glvalue core constant expression whose value refers to an object with static storage duration or to a function,

Upvotes: 1

Views: 68

Answers (2)

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171127

First off, you probably meant A * const i (a constant pointer to A) and not const A * i (a non-constant pointer to const A).

Still, even with this modification, it is perfectly legal to initialise a const variable with a value that is not a constant expression (such as a value computed at runtime). However, it is then not possible to use such a variable inside constant expressions. If you tried that, the constant expression definition would kick in and you'd get an error.

Upvotes: 4

filmor
filmor

Reputation: 32202

The initialisation of a const variable does not require a constant expression.

Upvotes: 0

Related Questions