if statement and --variable

Question

I am working toward my C++ exam, and I am doing some past papers. Then I stumbeled over this code understanding task. And I am wondering, why doesn't value change to 4 in this if statement?

int found = 0, value = 5;
if (!found || --value == 0) {
    cout << "danger";
}
cout << "value =" << value << endl;

You have read about [*short-circuit evaluation*](https://en.wikipedia.org/wiki/Short-circuit_evaluation)? — Some programmer dude, Apr 30 '18 at 10:23
Your first operation !found becomes true and its not executing --value . Learn about https://stackoverflow.com/questions/34492501/difference-between-and-or-and — Sumeet Kumar Yadav, Apr 30 '18 at 10:25
@BaumMitAugen: Why did you flip the meaning of the question? I know the question was incorrect before, but this could indicate a misunderstanding by OP. — interjay, Apr 30 '18 at 10:42
@interjay I just rolled back to the first, correct revision of the question. For some reason, OP edited it to make it wrong. Botched test or something I guess, dunno. — Baum mit Augen, Apr 30 '18 at 10:43

score 8 · Answer 1 · answered Apr 30 '18 at 10:28

8

if (!found || --value == 0)

In Logical OR operation, if first condition is satisfied, then it won't evaluate second condition. Here !found is TRUE and it won't execute --value thus leaving it's value as 5.

answered Apr 30 '18 at 10:28

Srikanth

131
8

score 0 · Answer 2 · answered Apr 30 '18 at 12:19

Because if the first condition is satisfied, the second is not even considered to evaluate. To get your expectations, swap the conditions.

int found = 0, value = 5;
if ( --value == 0 || !found) {
    cout << "danger";
}
cout << "value =" << value << endl;

if statement and --variable

2 Answers2