A constexpr function is not required to return a constant expression?

Question

C++ Primer (5th edition) on page 240 has a note stating:

\"A constexpr function is not required to return a constant ex

Answer 1

A constexpr function must return* must have a path that returns a constant expression iff all parameters are constant expressions. This actually makes sense. Example:

constexpr int square(int i){
    return i*i;
}

std::array<int, square(2)> ia; //works as intended, constant expression
int i;
std::cin >> i;
int j = square(i); //works even though i is not a constant expression
std::array<int, square(i)> ia; //fails, because square does not (and cannot)
                               //return a constant expression

*Correction by chris.

Answer 2

A (non-template) constexpr function must have at least one execution path that returns a constant expression; formally, there must exist argument values such that "an invocation of the function [...] could be an evaluated subexpression of a core constant expression" ([dcl.constexpr]/5). For example (ibid.):

constexpr int f(bool b) { return b ? throw 0 : 0; }     // OK
constexpr int f() { return f(true); }     // ill-formed, no diagnostic required

Here int f(bool) is allowed to be constexpr because its invocation with argument value false returns a constant expression.

It is possible to have a constexpr function that cannot ever return a constant expression if it is a specialization of a function template that could have at least one specialization that does return a constant expression. Again, with the above:

template<bool B> constexpr int g() { return f(B); }    // OK
constexpr int h() { return g<true>(); }    // ill-formed, no diagnostic required