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Clarification on behaviour of Logical OR in C?

+2 votes
419 views

I have been given an assignment to explain how the pieces of codes work. However, I have some doubts regarding it.

int i=-3,j=2,k=0,m;
m=++i && ++j && ++k;
printf("%d %d %d %d",i,j,k,m);

This piece of code returns -2 3 1 1, about which I have no confusion, but the problem arises when I try to run the second code:

int i=-3,j=2,k=0,m;
m=++i || ++j && ++k;
printf("%d %d %d %d",i,j,k,m);

I get -2 2 0 1 here, and how is that done? the ++j and ++k must return 3 and 1 respectively, according to the theory. Could anyone please explain me how it is executed?

posted Oct 13, 2014 by anonymous

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3 Answers

+1 vote
 
Best answer

op1 || op2
in case of || operator first op1 will be checked if it is non zero value op2 value not be checked or executed .
if op1 is zero then only next op2 will be checked .
So in your case op1 is ++j == -2 so op2 instruction will not be executed.
To verify this you can execute following instruction,
int i=-1,j=2,k=0,m;
m=++i || (++j && ++k);
printf("%d %d %d %d",i,j,k,m);

answer Oct 14, 2014 by Bheemappa G
+1 vote

My guess only (not tested)
if a||b and a is true is b is not executed so only ++i is executed and && is having higher precedence then || which means we can write the statement in the following way -
(++i) || (++j && ++k);

Others comment please

answer Oct 13, 2014 by Salil Agrawal
+1 vote

&& has higher precedence then || only in parse tree. But compiler optimizes the code as

if( !++i ) {
   ++j && ++k;
}

So here !++i is equal 0 in our case so, ++j && ++k is not evaluated.

answer Oct 14, 2014 by Arshad Khan
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