precedence of operators

[geekocoder]

can anyone please explain me the output of the following??
I am not able to figure out why arent the values of b and c not incremented???

#include <stdio.h>
int main()
{
int a=10,b=12,c=13;
int d=++a||++b&&++c;
printf("%d\t%d\t%d\t%d",a,b,c,d);
getchar();
return 0;
}

[Zhuge]

Try putting parentheses around every operator in the order the precedence table says they should be evaluated and see if that helps you figure out the expression.

[helios]

I am not able to figure out why arent the values of b and c not incremented???

It has nothing to do with precedence (syntax), and everything to do with semantics. Specifically, it's because of short-circuit evaluation.
Boolean OR (||) first evaluates its left hand operand (++a, in this case). If the operand is true (non-zero), the right hand operand (++b && ++c) is not evaluated, and the OR operation evaluates to true. Only if the left hand side is false is the right hand side evaluated.

This is because, if x is a boolean value,
(true || x) == true
(false || x) == x
Conversely,
(true && x) == x
(false && x) == false