将const分配给constexpr变量

Question

I tried to run a program based on constexpr .

Code:-

#include <iostream>
using namespace std;

int main()
{

        const int i = 10;
        constexpr int j = 10;

        constexpr int val1 = i;
        constexpr int val2 = j; 

        return 0;
}

In the book I follow, it is mentioned that if you assign a const to a constexpr variable, it is an error.

But my program compiles without any complaints.

Am I missing something?

Answer 1

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Ammendment

celtschk made a good point in the comments below the question. That is, you are not assigning to anything in your code. You are only initializing. Assigning from a const to a constexpr is indeed an error. So if that's what your book said, then it was not incorrect. However, this would be a strange point to make, since assigning in the other direction (from a constexpr to a const ) is also an error. Anyway, the rest of the answer is under the assumption that when you said "assign", you meant "initialize".

End of Ammendment

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Your book is incorrect (assuming you are not incorrectly paraphrasing what it said). A const integral which is initialized with a constant expression, is itself a constant expression. So i is a constant expression, and can be used to further initialize other constant expressions.

quoth the standard, 5.19/2

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:
...
— an lvalue-to-rvalue conversion (4.1) unless it is applied to:
...
— a non-volatile glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized with a constant expression
...

However, note that a const which is not initialized with a constant expression, is of course not a constant expression:

int a = 10;
const int b = a;
constexpr int c = b; // error

Also note that this only applies to integer and enum types. Not, for example floats and doubles.

const float a = 3.14;
constexpr float b = a; // error

Although some compilers might allow that (I believe MSVC does)