CODEWITHSUNDEEP
c
krishnang
krishnang

Reputation: 698

Output of the following C program

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.

Upvotes: 3

Views: 1741

Answers (2)

Azam Khan
Azam Khan

Reputation: 95

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.

Upvotes: 1

Jon
Jon

Reputation: 437734

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.

Upvotes: 12

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