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: http://stackoverflow.com/questions/14116003/difference-between-constexpr-and-const What is the purpose? Demonstration, I think. — user4581301, Apr 21 '17 at 05:42
Similar question, http://stackoverflow.com/questions/28614591/how-to-initialize-a-constexpr-reference. `constexpr` applies to the reference and `const int` is the thing referred to, they have different meanings — M.M, Apr 21 '17 at 05:58

ralismark · Answer 1 · 2017-04-22T06:52:37.330

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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 '17 at 06:52

answered Apr 21 '17 at 05:36

ralismark

"which means that the `const` is redundant" Hardly. – T.C. Apr 21 '17 at 06:34
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Unless memory fails me, since C++14 `constexpr` no longer implies `const`. – Andy Prowl Apr 21 '17 at 06:55
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@AndyProwl That's for member functions. – T.C. Apr 21 '17 at 07:08
A constexpr specifier used in an object declaration implies const ([constexpr specifier](http://en.cppreference.com/w/cpp/language/constexpr)) – ralismark Apr 21 '17 at 09:40
A reference is not an object. – Oktalist Apr 21 '17 at 19:25
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Try removing `const` from the code in the question. It won't compile. – Oktalist Apr 21 '17 at 19:36
@Oktalist, you are indeed right. I corrected the answer – ralismark Apr 22 '17 at 06:53
@Oktalist it does actually compile if you remove `const` – M.M Apr 22 '17 at 07:02
@Oktalist You also need to sub `3` with an lvalue but I dont think that is what OP was asking about (non-const lvalue reference cannot bind to rvalue, this is a separate issue to constexpr) – M.M Apr 22 '17 at 13:38
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@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. – Oktalist Apr 22 '17 at 14:46

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