Question

Why is k not getting incremented whereas,i and j are getting incremented in the same expression.And i also want to know what is the output of the program.I am getting the output as
-2 3 1 0

#include <stdio.h>
void main()
{
 int i=-3, j=2, m, k=0;
 m=++i && ++j || ++k;
 printf("%d %d %d %d", i, j, m, k);
}

Answer 1

The logical or, || short-circuits, and after

++i && ++j

the value of the entire expression is determined, so the right operand of the || isn’t evaluated.

m=++i && ++j || ++k;

is parenthesized m = (++i && ++j) || ++k; since the && has higher precedence than the ||.

The short-circuiting of the logical operators means that the right operand is only evaluated when the evaluation of the left has not yet determined the final result, for || that means the right operand is only evaluated if the left evaluated to 0, and for &&, the right operand is only evaluated if the left evaluated to a nonzero value.

So first ++i && ++j is evaluated, and for that, first ++i is evaluated. i had the value -3 before, so ++i evaluates to -2, which is not 0, hence the ++j is evaluated too. j had the value 2 before, so ++j evaluates to 3, which is again nonzero, and thus ++i && ++j evaluates to 1 (true). Since the left operand of the || is not zero, its result is already determined (to be 1), and the right operand isn’t evaluated, thus k remains unchanged and m is set to 1.