以下C程序的行为是否未定义？

Question

I'm looking in one of my old courses and I find the following:

int main(void) {
    int x = 0;
    return (x = 1) + (x = 2);
}

"According to C standard, the behavior of the program above is undefined.

GCC4, MSVC: returns 4

GCC3, ICC, Clang: returns 3"

There's a slide after this saying the following code is not undefined in the C standard. Can anyone explain to me why it's not undefined behavior?

int main(void) {
    int x = 0;
    int y = 2;
    return (x = 4) + (x=(x + y)/2);
}

Answer 1

Relevant passages from the C specification:

Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B. (Conversely, if A is sequenced before B, then B is sequenced after A.) If A is not sequenced before or after B, then A and B are unsequenced.

and

If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. If there are multiple allowable orderings of the subexpressions of an expression, the behavior is undefined if such an unsequenced side effect occurs in any of the orderings.

---

Thus, both snippets in your question invoke undefined behavior.

(Because originally your second snippet was C++, it's worth mentioning that even though recent versions of C++ have expanded the notion of what "sequenced before" and "sequenced after" mean, AFAIK it still doesn't include operator+ as a sequence point.)