operators - Short-circuit evaluation on C

Question

I'm studying C from A Book on C by Kelley-Pohl, and there's this exercise that I don't understand:

int a = 0, b = 0, x;

x = 0 && (a = b = 777);
printf("%d %d %d
", a, b, x);
x = 777 || (a = ++b);
printf("%d %d %d
", a, b, x);

They just say to imagine the output and compare it to the real one. I thought the output would have been

777 777 0

778 778 1

but it is

0 0 0

0 0 1

Answer 1

The && operator uses lazy evaluation. If either side of the && operator is false, then the whole expression is false.

C checks the truth value of the left hand side of the operator, which in your case is 0. Since 0 is false in c, then the right hand side expression of the operation, (a = b = 777), is never evaluated.

The second case is similar, except that || returns true if the left hand side expression returns true. Also remember that in c, anything that is not 0 is considered true.

Hope this helps.