As far as I can tell the function below is not constexpr, but the code compiles in clang and g++. What am I missing?

Question

I got this example from §5.19/2 in N4140:

constexpr int incr(int &n) {
    return ++n;
}

As far as I can tell, this is not a constexp

Answer 1

As far as I can tell, this is not a constexpr function.

Why do you say that? The example from §5.19/2 reads:

constexpr int g(int k) {
    constexpr int x = incr(k); // error: incr(k) is not a core constant
                               // expression because lifetime of k
                               // began outside the expression incr(k)
    return x;
}

incr(k) not being a core constant expression does not mean incr cannot not be a constexpr function.

Under C++14's constexpr rules, it is possible to use incr in a constexpr context, for example:

constexpr int incr(int& n) {
    return ++n;
}

constexpr int foo() {
    int n = 0;
    incr(n);
    return n;
}

Unless it's downright impossible for the body of the function to be constexpr (for example, calling a non-constexpr function unconditionally), the compiler has no reason to produce an error at the point of definition.

A constexpr function may even contain paths/branches in the body which would not be constexpr. As long as they are never taken in a constexpr context, you will not get an error. For example:

constexpr int maybe_constexpr(bool choice, const int& a, const int& b) {
    return choice ? a : b;
}

constexpr int a = 0;
int b = 1;
static_assert(maybe_constexpr(true, a, b) == 0, "!");

live example