Unexpected behavior of complicated expression with increments and logical operators

Question

I was going through a C objective book where a question comes:

#include<stdio.h>
#include<conio.h> 
int main()
{
    int i,j,k;
    i=j=k=1;
    k=++i||++j&&++k;
    printf("%d %d %d",i,j,k);
    return 0;
}

The output is:

2 1 1

In my view:

So the output should be i=2,j=2,k=1. What am I missing?

Carl Norum · Accepted Answer · 2013-09-11T20:55:21.823

3

The expression k=++i||++j&&++k; causes undefined behaviour. You are trying to assign to k twice without an intervening sequence point.

Even if the assignment were to a different variable, your steps would be inaccurate - the logical operators have short-circuiting behaviour.

Edit: OP says he changed the expression to a=++i||++j&&++k. I'm going to rewrite it fully parenthesized and with some spaces:

a = ++i || (++j && ++k);

In this case, only the ++i is evaluated, due to short-circuiting behaviour of the || operator.

edited Sep 11 '13 at 20:55

answered Sep 11 '13 at 20:47

Carl Norum

4

(Also, if a programming book invokes undefined behavior, and doesn't state that it's undefined behavior, you should probably get a different book) – Dennis Meng Sep 11 '13 at 20:48
was my logic wrong if we don't go to undefined behaviour.. – mrigendra Sep 11 '13 at 20:50
@migrenda, yes, your logic is still wrong, even in the absence of the UB in this case. – Carl Norum Sep 11 '13 at 20:51
I changed k to a(a=++i||++j&&++k) – mrigendra Sep 11 '13 at 20:52
out put is still 2 1 1 – mrigendra Sep 11 '13 at 20:52
OK, so what's the problem? `++j&&++k` is never executed due to short-circuiting. – Carl Norum Sep 11 '13 at 20:53
i have checked the precedence table and postfix is given higher priority in left to right sequence so maybe the leftmost (i.e k) will be evaluated first. According to you ++i will be done, then OR will be skipped and then AND operation will be done. – mrigendra Sep 11 '13 at 20:57
I think you'd better read more about how operator precedence and short-circuiting works in C. – Carl Norum Sep 11 '13 at 20:57
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/37209/discussion-between-mrigendra-and-carl-norum) – mrigendra Sep 11 '13 at 21:01

1 Answers1