CODEWITHSUNDEEP
java
Lego_blocks
Lego_blocks

Reputation: 301

cannot add one to variable pass by a function

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;
    }
}

Upvotes: 0

Views: 80

Answers (6)

Jens
Jens

Reputation: 69470

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){

Upvotes: 1

Ninad Pingale
Ninad Pingale

Reputation: 7079

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.

Upvotes: 1

tudoricc
tudoricc

Reputation: 729

This is the main difference between Java and C for example.In your case you don't send the exact item x you are jsut sending a "copy" of it to the function , so the result of the function doesn't modify x's adress just the local value of the copied object x.

The most common way of solving this is just assigning the result of the function to x

 x=OnePlusNumber(x);

Upvotes: 0

Sagar Gandhi
Sagar Gandhi

Reputation: 965

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

also read about

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

Upvotes: 0

Iffo
Iffo

Reputation: 353

The function OnePlusNumber(x); return a Integer.

Replace it with x = OnePlusNumber(x);

Upvotes: 0

Stultuske
Stultuske

Reputation: 9437

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

 OnePlusNumber(x);

to

x = OnePlusNumber(x);

Upvotes: 3

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