Short circuit evaluation and side effects

Question

OK, I'm a little embarassed to ask this question, but I just want to be sure...

It is known that C uses short circuit evaluation in boolean expressions:

int c = 0;
if (c && func(c)) { /* whatever... */ }

In that example func(c) is not called because c evaluates to 0. But how about more sophisticated example where side effects of comparison would change the variable being compared next? Like this:

int c; /* this is not even initialized... */
if (canInitWithSomeValue(&c) && c == SOMETHING) { /*...*/ }

Function canInitWithSomeValue returns true and changes value at given pointer in case of success. Is it guaranteed that subsequent comparisons (c == SOMETHING in this example) uses value set by canInitWithSomeValue(&c)?

No matter how heavy optimizations the compiler uses?

Answer 1

Is it guaranteed that subsequent comparisons (c == SOMETHING in this example) uses value set by canInitWithSomeValue(&c)?

Yes. Because there is a sequence point

Between evaluation of the left and right operands of the && (logical AND), || (logical OR), and comma operators. For example, in the expression *p++ != 0 && *q++ != 0, all side effects of the sub-expression *p++ != 0 are completed before any attempt to access q.

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A sequence point defines any point in a computer program's execution at which it is guaranteed that all side effects of previous evaluations will have been performed, and no side effects from subsequent evaluations have yet been performed.

Answer 2

Yes. Because both && and || operator are also something called sequence points. The latter define when the side-effects of a previous operation should be complete and those of the next should not have begun.

Answer 3

Evaluation within the if statement composite condition is strictly left to right. The only circumstance under which the second test in your if would be optimized out is if the compiler can determine with 100% certainty that the first is identically equal to false.