Initializing constexpr with const: Different treatment for int and double [duplicate]

Answer 1

Rule:"constexpr must be evaluate at compile time".

Let's look below code (generic example);

    const double k1 = size_of_array(); 

k1 is constant, the value of its initializer is not known compile time but its initializer is known until run time so k1 is not constant expression. As a result a const variable is not constexpr.

But compiler see these code:

    const int k1 = 10;  
    constexpr int k2 = 2*k1;

One exception occurs. A constexpr integral value can be used wherever a const integer is required, such as in template arguments and array declarations [1].

You can get extra information from the links below:

1. Constexpr - Generalized Constant Expressions in C++11
2. const vs constexpr on variables | stackoverflow
3. Difference between constexpr and const | stackoverflow