Relation between constexpr and pure functions

Question

Am I right, that:

• Any function defined with constexpr is a pure function, and
• Any pure function can be and must be defined with constexpr if it's not very expensive for compiler.

And if so, why arent <cmath>'s functions defined with constexpr ?

Answer 1

To add to what others have said, consider the following constexpr function template:

template <typename T> constexpr T add(T x, T y) { return x + y; } 

This constexpr function template is usable in a constant expression in some cases (e.g., where T is int) but not in others (e.g., where T is a class type with an operator+ overload that is not declared constexpr).

constexpr does not mean that the function is always usable in a constant expression, it means that the function may be usable in a constant expression.

(There are similar examples involving nontemplate functions.)

Answer 2

In addition to the previous answers: constexpr on a function restricts its implementation greatly: its body must be visible to the compiler (inline), and must consist only of a single return statement. I'd be surprised if you could implement sqrt() or sin() correctly and still meet that last condition.