CODEWITHSUNDEEP
javaoperator-keyword
D. Ace
D. Ace

Reputation: 418

+ operator - Java pass by value

I can't figure out why line 3 prints 1 in this Test class that I wrote. I thought it would print 2 instead of 1 because I was thinking line 2 has changed the value of x to 2.

Can anyone explain this to me please? Is this to do with java being pass by value or is it some other explanation? 

public class Test{

    public static void main(String[] args) {
        int x = 1;  // line 1
        System.out.println(x+1); //line 2 prints 2
        System.out.println(x); //line 3 but why does this line prints 1?

    }

}

output:

2

1

Upvotes: 0

Views: 49

Answers (3)

Ravindra babu
Ravindra babu

Reputation: 38910

It is expected.

You are just printing x+1. x+1 does not modify the value of x unless you explicitly assign the new value to x.

if you change the code as below, you will get desired output.

x= x+1;
System.out.println(x);

Upvotes: 0

Osama Aftab
Osama Aftab

Reputation: 1181

For Line 2 the output is 2 because x is already 1 and you're adding an other one that makes it 2 but not changing the value of x. So that is the reason for 1 on line 3 aswell.

Try with x++ on line 2 and see the result on line three.

Upvotes: 2

Andy Turner
Andy Turner

Reputation: 140299

The explanation is simply that x + 1 doesn't change the value of x, but instead returns a completely separate int whose value is one greater than x.

Upvotes: 3

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