How does this code produces an output of 6 and 3?

Question

#include<stdio.h>
void main() {
     int x = 0,y = 0,k = 0;
     for(k = 0; k < 5; k++){
         if(++x > 2 && ++y > 2)    x++;
     }
     printf("x = %d and y = %d",x,y);
}

I'm not able to understand how the above piece of code generates x = 6 and y = 3.

RESOLVED : I didn't know that when there is &&, if the first statement evaluates to false the second will not be executed.

What don't you understand about it? The increments? The if statement? What? — , May 22 '17 at 13:02
Use a debugger or `printf`to see the value of each variable for each iteration. It shall help you understand how it works step by step — Badda, May 22 '17 at 13:04
I retracted my vote.....don't know why it wasn't. I voted to reopen — LPs, May 22 '17 at 13:15
I re-opened the post. Although... possible duplicate: https://stackoverflow.com/questions/628526/is-short-circuiting-logical-operators-mandated-and-evaluation-order — Lundin, May 22 '17 at 13:34

pmg · Accepted Answer · 2017-05-22T13:11:39.127

4

&& is a short-circuit operator.

The first time through the loop, only ++x is evaluated.
The second time through the loop, only ++x is evaluated.
The third time through the loop, both are evaluated.
...

edited May 22 '17 at 13:11

answered May 22 '17 at 13:04

pmg

106,608
13
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gsamaras · Answer 2 · 2017-05-22T13:06:46.133

Not related to your question, but please read What should main() return in C and C++? int.

c enables short circuit, and && is an operator that follows that. So, this:

if(++x > 2 && ++y > 2)

says:

Increment x by 1.
If x is greater than 2 (thus the first operand of && is true), evaluate the second operand.
The second operand says to increment y by 1, and if y > 2 is true, then the whole if condition will be true.

Your code is equivalent to this:

#include <stdio.h>
int main() {
     int x = 0, y = 0, k = 0;
     for(k = 0; k < 5; k++){
         x = x + 1;
         if(x > 2)
         {
             y = y + 1;
             if(y > 2)
             {
                 x = x + 1;
             }
         }  
     }
     printf("x = %d and y = %d", x, y);
     return 0;
}

keep_smiling · Answer 3 · 2017-05-22T14:41:52.087

&& is the short-circuit operator.

if ( ++x > 2 && ++y > 2 )

in this if statement the second operand will be evaluated only if the first one is true.

when k=0 X will be incremented by 1. Now x value is 1. x > 2 is false. So Y won't increase.
When k=1 X will be incremented by 1 . Now X value is 2 . X > 2 is false. So Y won't increase.
when k=2 X will be incremented by 1 . Now X value is 3 . X > 2 is true . So Y will be incremented by 1 . Now Y value is 1 . but Y > 2 is false . So total if condition is false.
when k=3 X will be incremented by 1 . Now X value is 4 . X > 2 is true . So Y will be incremented by 1 . Now Y value is 2 . but Y > 2 is false . So total if condition is false.
when k=4 X will be incremented by 1 . Now X value is 5 . X > 2 is true . So Y will be incremented by 1 . Now Y value is 3 . Y > 2 is true . So total if condition is true. Then X will be incremented by 1.
The final answer is X=6 and Y=3 .

I don't like how this is worded: "the first operand is true then only the second operand will be evaluated." It looks like you are saying that the second operand will be evaluated and the first one won't. I'm not a native speaker, but I think you should change it to "only if the first operand is true will the second one be evaluated" to avoid the ambiguity (to any native speakers reading this, can you please let me know whether I am right?). Or, to keep it simple, I would rephrase it as "in this `if` statement the second operand will be evaluated only if the first one is true." — Fabio says Reinstate Monica, May 22 '17 at 14:38

score -1 · Answer 4 · answered May 22 '17 at 13:06

-1

if(++x > 2 && ++y > 2)

In this line if first condition is false it will not evalute second condition. So first condition is false until value of x is 3

answered May 22 '17 at 13:06

Sunil Hirole

239
3
11

How does this code produces an output of 6 and 3?

4 Answers4