cannot add one to variable pass by a function

Question

I cannot add one to the integer on the function below, it still prints 5. Can anyone explain this?

public class HelloWorld {
    public static void main(String[] args) {
        int x = 5;
        System.out.print('Hello world~~~~~');
        for(int i = 0; i < args.length; i++) { 
            System.out.println(args[i]);
        }
        System.out.println(Runtime.getRuntime().maxMemory());
        System.out.println(Runtime.getRuntime().totalMemory());
        System.out.println(Runtime.getRuntime().freeMemory());
        OnePlusNumber(x);
        System.out.println(x);      

        Date date = new Date();
        System.out.println(date);       
    }

    private static Integer OnePlusNumber(int number) {
        number += 1;        
        return number;
    }
}

Java is always pass-by-value, so inside your method you edit a cpoy of x, not x itself (more here: http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value ) — kajacx, Jul 09 '14 at 11:18

score 3 · Accepted Answer · answered Jul 09 '14 at 11:17

3

you don't assign the value you return. change the following line:

 OnePlusNumber(x);

to

x = OnePlusNumber(x);

answered Jul 09 '14 at 11:17

Stultuske

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Ninad Pingale · Answer 2 · 2014-07-09T11:47:41.330

1

Change

OnePlusNumber(x);

to

x=OnePlusNumber(x);

It will assign returned value from the method to x variable again, as it is a primitive data type (int).

If the passed parameter would have been an object of a class, you did not have to assign it like this. As same object's state gets changed when reference is passed to a method -

for ex-

Employee emp=new Employee();
emp.setName("A");

changeEmpName(emp);


public void changeEmpName(Employee employee){
 employee.setName("B");
}

Then employees name becomes B.

This method will change original emp object , as it's reference was passed.

edited Jul 09 '14 at 11:47

answered Jul 09 '14 at 11:18

Ninad Pingale

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – stealthjong Jul 09 '14 at 11:38
Doesn't it a solution for the problem ? should I delete the answer ? – Ninad Pingale Jul 09 '14 at 11:40
@stealthjong: Thanks added some explaination – Ninad Pingale Jul 09 '14 at 11:49

score 1 · Answer 3 · edited Dec 11 '15 at 23:22

1

You have to change yout code:

OnePlusNumber(x);

Should be

x = OnePlusNumber(x);

So you have the return value.

And your method schould reurn an int:

  private static int OnePlusNumber(int number){

edited Dec 11 '15 at 23:22

Josh Crozier

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answered Jul 09 '14 at 11:18

Jens

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score 0 · Answer 4 · answered Jul 09 '14 at 11:19

0

The function OnePlusNumber(x); return a Integer.

Replace it with x = OnePlusNumber(x);

answered Jul 09 '14 at 11:19

Iffo

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score 0 · Answer 5 · answered Jul 09 '14 at 11:22

Java uses reference. When u pass x to OnePlusNumber. It passes reference of x, but in java , primitive types and string are immutable. X is Integer here. so number+=1 will create new Integer. but X in original function , refers to old x.

That's why you need to assign return value.

x = OnePlusNumber(x);

http://en.wikipedia.org/wiki/Primitive_wrapper_class

http://docs.oracle.com/javase/tutorial/java/data/numberclasses.html

cannot add one to variable pass by a function

6 Answers6