Calling function with side effects inside expression

Question

I thought I understand how sequence points work in C++, but this GeeksQuiz question puzzled me:

int f(int &x, int c) {
    c = c - 1;
    if (c == 0) return 1;
    x = x + 1;
    return f(x, c) * x;
}

int main() {
    int p = 5;
    cout << f(p, p) << endl;
    return 0;
}

The “correct” answer to this question says it prints 6561. Indeed, in VS2013 it does. But isn't it UB anyway because there is no guarantee which will be evaluated first: f(x, c) or x. We get 6561 if f(x, c) is evaluated first: the whole thing turns into five recursive calls: the first four (c = 5, 4, 3, 2) continue on, the last one (c = 1) terminates and returns 1, which amounts to 9 ** 4 in the end.

However, if x was evaluated first, then we'd get 6 * 7 * 8 * 9 * 1 instead. The funny thing is, in VS2013 even replacing f(x, c) * x with x * f(x, c) doesn't change the result. Not that it means anything.

According to the standard, is this UB or not? If not, why?

Answer 1

This is UB.

n4140 §1.9 [intro.execution]/15

Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [...] If a side effect on a scalar object is unsequenced relative to [...] value computation using the value of the same scalar object [...] the behavior is undefined.

Multiplicative operators don't have sequencing explicitly noted.