Use of constexpr and const in reference definition

Question

I have seen declaring a reference variable as constant in C++ on Quora.

static constexpr const int& r = 3;

So, Why both constexpr and const used in single statement?

What is the purpose of that type of statement?

Useful reading: stackoverflow.com/questions/14116003/… What is the purpose? Demonstration, I think. — user4581301
– user4581301, Commented Apr 21, 2017 at 5:42
Similar question, stackoverflow.com/questions/28614591/…. constexpr applies to the reference and const int is the thing referred to, they have different meanings — M.M
– M.M, Commented Apr 21, 2017 at 5:58

ralismark · Accepted Answer · 2017-04-22 06:52:37Z

6

const variables are ones that cannot be modified after initialisation (e.g. const int a = 1).

constexpr variables are constant expressions and can be used at compile-time. The use of constexpr for a variable declaration implies const.

However, in this declaration, const applies to the int, while constexpr applies to const int& (a reference to a const int).

edited Apr 22, 2017 at 6:52

answered Apr 21, 2017 at 5:36

ralismark

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10 Comments

T.C. Over a year ago

"which means that the const is redundant" Hardly.

Andy Prowl Over a year ago

Unless memory fails me, since C++14 constexpr no longer implies const.

T.C. Over a year ago

@AndyProwl That's for member functions.

Oktalist Over a year ago

Try removing const from the code in the question. It won't compile.

Oktalist Over a year ago

@M.M Obviously you can drop the const if you also change the initializer. My point stands: the answer was wrong to say the const was redundant, as removing the const changed the meaning.

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