Increamenting odd values using while loop shows wrong required values

Question

Why the attached code gives me this out put :

3 , 5 , 7 , 9

it should show the even values only

public static void main (String [] args){
    int i = 1;
    
    while ( i < 10 ){
        if ((i++)%2 == 0){
            System.out.println(i);
          
        }
    }}

You want to increment `i` after printing the value: `if(i%2 == 0){System.out.println(i++);}` — Mady Daby, Apr 21 '21 at 19:27
You are checking the condition with i and then print i+1 because of post-increment. — Iman H, Apr 21 '21 at 19:28

score 0 · Answer 1 · answered Apr 21 '21 at 19:34

0

This is because you are doing i++ inside the if. The post increment there means that the value evaluated in the condition is pre-increment but when you reach the print statement, it has been incremented. Try incrementing after the if.

answered Apr 21 '21 at 19:34

codetective

86
2
9

score 0 · Answer 2 · answered Apr 21 '21 at 19:44

0

Try this code:

public static void main (String [] args){
int i = 1;

while ( i < 10 ){
    if ((++i)%2 == 0){
        System.out.println(i);
      
    }
}}

answered Apr 21 '21 at 19:44

nissim abehcera

821
6
7

score 0 · Answer 3 · answered Apr 21 '21 at 19:51

The evaluation is done prior to the increment.
the printing is done after the increment.

int i = 1;

while ( i < 10 ){
    if ((i++)%2 == 0)  {        // i evaluated true while "even" before increment
        System.out.println(i);  // printing post increment i 
    }
}

To correct this, simply move the autoincrement to the left of i.

Increamenting odd values using while loop shows wrong required values

3 Answers3