constexpr vs const vs constexpr const

Question

const-vs-constexpr-on-variables

What the guy says about constexpr is right if double is used (or float of course). However, if you change t

Answer 1

Based on your answer in the comment this is my answer. The C++ standard makes it pretty clear. GCC 5.1 works pretty fine here though: https://godbolt.org/g/2oV6Hk

A converted constant expression of type T is an expression, implicitly converted to type T, where the converted expression is a constant expression and the implicit conversion sequence contains only § 5.20 134 c ISO/IEC N4567

[...]

(4.6) — integral promotions (4.5),

(4.7) — integral conversions (4.7) other than narrowing conversions (8.5.4),

[...]

For the reference for narrowing conversions (8.5.4/7) in n4567:

A narrowing conversion is an implicit conversion

• from a floating-point type to an integer type, or
• from long double to double or float, or from double to float, except where the source is a constant expression and the actual value after conversion is within the range of values that can be represented (even if it cannot be represented exactly), or
• from an integer type or unscoped enumeration type to a floating-point type, except where the source is a constant expression and the actual value after conversion will fit into the target type and will produce the original value when converted back to the original type, or
• from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression and the actual value after conversion will fit into the target type and will produce the original value when converted back to the original type.

Answer 2

The compiler does not allow an implicit narrowing or non-integral promotion during the initialisation of a constexpr variable.

This will work:

int main()
{
    const int PI1 = 3;
    constexpr int PI2 = 3;
    constexpr int PI3 = PI1;  // works
    static_assert(PI1 == 3, "");  // works

    const double PI1__ = 3;
    constexpr double PI2__ = 3;
    constexpr double PI3__ = double(PI1);  // works with explicit cast
    static_assert(PI2__ == 3, "");  // works now. PI1__ isn't constexpr
    return 0;
}

Answer 3

From the comments it seems like OP is asking for Standard quote which defines const int as a compile-time constant, but const double as not.

The corresponding details are found in 5.19, Constant Expressions. In particular:

...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...

int is an integral type, while double is not.