Is this undefined?

Question

My compiler warns: operation on j may be undefined

Here's the C code:

for(int j = 1; pattern[j] != '\0' && string[i] != '\0';){
    if(string[i+j] != pattern[j++]){//this is on the warning
        found = 0;
        break;
    }
}

Is that undefined?

You might find [this question](http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points?rq=1) an interesting read. — WhozCraig, Sep 02 '13 at 11:27
I think problem is related to [Sequence points](http://www.geeksforgeeks.org/sequence-points-in-c-set-1/) — 0xF1, Sep 02 '13 at 11:27
Usually when someone asks whether something is UB in C, it is. — harold, Sep 02 '13 at 11:45
possible duplicate of [Why are these constructs undefined behavior?](http://stackoverflow.com/questions/949433/why-are-these-constructs-undefined-behavior) — Pascal Cuoq, Sep 27 '13 at 14:07

Dayal rai · Accepted Answer · 2013-09-27T13:43:18.197

10

YES. string[i+j] != pattern[j++] is having two different execution based on variable j without any sequence point in between. So it is example of undefined behaviour.

edited Sep 27 '13 at 13:43

answered Sep 02 '13 at 11:26

Dayal rai

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The [page](http://www.geeksforgeeks.org/sequence-points-in-c-set-1/) you requested is _now available_ :-) – 0xF1 Sep 02 '13 at 11:31

score 2 · Answer 2 · answered Sep 02 '13 at 11:39

Yes. The C11 standard says in §6.5:

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.

Here, in the comparison

if(string[i+j] != pattern[j++])

you are both incrementing the value of j with pattern [j++], and using the value of j in string [i + j]. The side effect of j++ is not sequenced relative to the value computation i + j. So this is classic undefined behavior.

Is this undefined?

2 Answers2