Why constexpr method can return correctly class members whose value change during execution?

Question

I just discovered that a constexpr method can return correctly the value of a class member that changes during execution. My question is, how is this possible if constexpr methods are supposed to be evaluated completely at compile time?

The example below correctly outputs Value: 0 and then Value: 5. Even more, if I change the a.change(5) to something unpredictible for the compiler (such as a.change(atoi(argv[1])) or a.change(rand() % 10 + 1) it still works. Why? Why does it even compile?

#include <iostream>

class A
{
    public:
        void change(int value) { x = value; }
        int get() const { return x; }
        constexpr int value() const noexcept { return x; }
    private:
        int x{0};
};

int main()
{
    A a;

    std::cout << "Value: "  << a.get() << std::endl;
    a.change(5);
    std::cout << "Value: "  << a.get() << std::endl;
}

Thank you in advance

Answer 1

In your example you're not even calling the constexpr function, you're calling get, which is not constexpr.

However, constexpr functions can be evaluated at compile time, but also during run time if compile time evaluation is not possible.

If you called value instead of get, it would still work and the function would be evaluated at run time.

In order to force compile time evaluation, you can write

constexpr auto val = a.value();

This will indeed give you an error if compile time evaluation is not possible.