Basic programming exercise about logical operators

Question

I have a problem with a question in my book:

#include<stdio.h>
void main()
{
    int a=5, b=-7, c=0, d;
    d = ++a && ++b || ++c;
    printf("\n%d%d%d%d",a,b,c,d);
}

The question asks me what is the output of the code. I ran it and the result on the screen is 6-601. I understand why a=6 and b=-6, but I don't understand why c=0 and d=1?

Your answer got accepted. That's about as much proof as you would need. I sometimes wonder if I'm not overdoing it, when I write an answer like that, and by the time I'm done, someone has written a 'Try this ``' answer that got accepted instead. — GolezTrol, Dec 29 '15 at 07:03
@GolezTrol I think sir you meant to post this comment under my answer, however, thank you. :) — Sourav Ghosh, Dec 30 '15 at 20:36
@SouravGhosh Ah, that's where that comment went! I meant to post it under [your question on Meta](http://meta.stackoverflow.com/questions/313420/how-much-explanation-is-good-am-i-overdoing-it), but then I couldn't find it anymore. :) — GolezTrol, Dec 30 '15 at 21:07

Sourav Ghosh · Accepted Answer · 2015-12-25T06:32:03.020

I believe you already got your answer, but just to elaborate a bit step-by-step, let me add one more clarification here. Firstly, to quote the properties of the && and || operators, from C11 standard, chapter §6.5.13 and §6.5.13, respectively,

(I)

The && operator shall yield 1 if both of its operands compare unequal to 0; otherwise, it yields 0. [...] If the first operand compares equal to 0, the second operand is not evaluated.

and

(II)

The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. [...]. If the first operand compares unequal to 0, the second operand is not evaluated.

and they both guarantee left-to-right evaluation. So, comparing your code,

d = ++a && ++b || ++c;

it happens like

d = ((++a && ++b) || ++c );

which evaluates to

d = (( 6 && ++b ) || ++c);

and then

d = ( ( 6 && (-6) ) || ++c);

Now in above stage, (I) is fulfilled and it comes down to

d = ( 1 || ++c);

Now, following the emphasis, which already meets the (II), so no further evaluation of the RHS operand of || is performed (i.e., ++c is not evaluated), and it appears to be d = 1 and the final result, 1, is stored into d.

That's how, a == 6, b == -6, c == 0 and d ==1.

Having said that, void main() should be changed to int main(void), at least to conform with the standard.

Thank you, your explanation is very clear, but i'm confused between void main() and int main(void), the code above i just copied directly from my book. — Hoang.T, Dec 25 '15 at 07:15
@Hoang.T there is very good discussion posted [here](http://stackoverflow.com/q/204476/2173917). Do read it. :) — Sourav Ghosh, Dec 25 '15 at 07:17

score 14 · Answer 2 · answered Dec 25 '15 at 06:01

The || OR operator is short-circuiting, which means that if the left side is true then the right side is not evaluated. In this case ++a && ++b evaluates to true, so ++c is never run and c keeps its value of zero.

Also since it evaluates to true, this is denoted with 1 which is stored in d.

Any non-zero value is considered to be true and the result of boolean operations is defined to be 0 or 1 as an integer.

Basic programming exercise about logical operators

2 Answers2