Output of the following C program

Question

What should be the output of this C program?

#include<stdio.h>
int main(){
  int x,y,z;
  x=y=z=1;
  z = ++x || ++y && ++z;
  printf("x=%d y=%d z=%d\n",x,y,z);
  return 0;
}

The given output is : x=2 y=1 z=1
I understand the output for x, but fail to see how y and z values don't get incremented.

Possible duplicate of [Why does "++x || ++y && ++z" calculate "++x" first, even though operator "&&" has higher precedence than "||"](https://stackoverflow.com/questions/3700352/why-does-x-y-z-calculate-x-first-even-though-operator-ha) — phuclv, Aug 18 '18 at 11:19

Jon · Accepted Answer · 2012-05-03T14:05:31.753

This is a result of short-circuit evaluation.

The expression ++x evaluates to 2, and the compiler knows that 2 || anything always evaluates to 1 ("true") no matter what anything is. Therefore it does not proceed to evaluate anything and the values of y and z do not change.

If you try with

x=-1;
y=z=1;

You will see that y and z will be incremented, because the compiler has to evaluate the right hand side of the OR to determine the result of the expression.

Edit: asaerl answered your follow-up question in the comments first so I 'll just expand on his correct answer a little.

Operator precedence determines how the parts that make up an expression bind together. Because AND has higher precedence than OR, the compiler knows that you wrote

++x || (++y && ++z)

instead of

(++x || ++y) && ++z

This leaves it tasked to do an OR between ++x and ++y && ++z. At this point it would normally be free to select if it would "prefer" to evaluate one or the other expression first -- as per the standard -- and you would not normally be able to depend on the specific order. This order has nothing to do with operator precedence.

However, specifically for || and && the standard demands that evaluation will always proceed from left to right so that short-circuiting can work and developers can depend on the rhs expression not being evaluated if the result of evaluating the lhs tells.

Thanks that helped, but doesn't the order of precedence come into play here? Meaning the ++y && ++z should be evaluated first, before the || ? — krishnang, May 03 '12 at 13:52
No. precedence is order in parsing, not order of evaluating. BTW, if `++z` was been evaluated, it was UB. (changing `z` twice) — asaelr, May 03 '12 at 13:57

score 1 · Answer 2 · answered Sep 11 '14 at 13:49

In C, any thing other than 0 is treated as true, And the evaluation for the || start from left to right.

Hence the compiler will check first left operand and if it is true then the compiler will not check other operands. ex. A || B - In this case if A is true then compiler will return true only, and will not check whether B is true or False. But if A is false then it will check B and return accordingly means if B is true then will return true or if B is false then it will return false.

In your program compiler first will check ++x(i.e 2) and anything other than 0 is true in C. Hence it will not check/increment other expressions.

Output of the following C program

2 Answers2