Confused about memory in java

Question

From my understand objects in java are passed by reference, or to be more exact, the references to objects are passed by value. So, if I declare a string and pass it into a function where I change the value of the string, why doesn't the original string change? For example:

class Thing 
{

    static void func(String x){ x = "new"; }

    public static void main(String [] args)
    {
        String y = "old";
        func(y);
        System.out.print(y);
    }
}

Why is the value of y still "old"?

EDIT: Why does the following set the attribute x of Thing something to 90?

class Thing { int x = 0; 

static void func(Thing t){ t.x = 90; }

public static void main(String [] args){
    Thing something = null;
    something = new Thing();
    func(something);
    System.out.print(something.x);
}
}

score 2 · Accepted Answer · edited May 23 '17 at 10:31

First of all everything in java is pass by value. Even references are also pass by value.

You have created a new string literal and not returning that in the method func().

You are modifying the passed argument not the original string, hence you can't see the changes.

You might need this,

public static void main(String [] args)
    {
        String y = "old";
        y= func(y);
        System.out.print(y);
    }

 static String func(String x){ 
  x = "new";  
  return x
 }

Edit for the comment:

No they both are not Identical. There a lot of difference between two ways of String, especially in the memory aspect.

Read

Edit2:

In first case you are created a new string literal in the func method, but here you are modifying the reference.

Your doubt clarifies when you done

  static void func(Thing t){
     t = new Thing(); //as like previous example
     t.x = 90;       // points to new one. Not the original.

    }

Now check the result.

why does the following set the attribute x of Thing to 90? (See edit in question) — SKLAK, Oct 19 '13 at 06:32
@SKLAK Please check the update. You both examples are different. In first one you creating a new string and in second case you are not creating. — Suresh Atta, Oct 19 '13 at 06:49

score 0 · Answer 2 · edited Oct 19 '13 at 16:36

0

As stated here:

Java passes everything by value, and not by reference – make sure you remember that. And when we say everything, we mean everything – objects, arrays (which are objects in Java), primitive types (like ints and floats), etc. – these are all passed by value in Java. What is the difference between pass by value and pass by reference? When passing an argument (or even multiple arguments) to a method, Java will create a copy or copies of the values inside the original variable(s) and pass that to the method as arguments – and that is why it is called pass by value. The key with pass by value is that the method will not receive the actual variable that is being passed – but just a copy of the value being stored inside the variable.

edited Oct 19 '13 at 16:36

Brad Larson

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answered Oct 19 '13 at 06:11

AJ.

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Then why are we able to change attributes of an object. For example say you have a class Car which has attributes speed and color. Say you have a static function in this class which sets the speed to the passed in car to 100. If I create a Car called carx inside main and do something like setSpeed(carx), won't carx.getSpeed() return 100? – SKLAK Oct 19 '13 at 06:29
1

You can't just dump a link at the bottom of an answer copied from somewhere else. You need to clearly quote this content to show that none of these are your original words. I've done this in the above. – Brad Larson Oct 19 '13 at 16:37
"we mean everything – objects, arrays (which are objects in Java), primitive types (like ints and floats), etc. – these are all passed by value in Java" "Objects" are not values in Java and cannot be "passed". – newacct Oct 20 '13 at 22:38

Vaibhav Raj · Answer 3 · 2013-10-19T06:27:50.963

It has nothing to do with immutability. It's all about how parameters are passed and the scope of variables.

Here is your function call

String y = "old"; // The reference y pointing to the string object "old": y ---> "old"
func(y);

and when you enter the called function

static void func(String x){  // x = y which means x now points to "old": x ---> "old"
    x = "new"; // x ---> "new": now the local variable x is pointing to a different object. y still points to "old".
} //Scope of x ends here

When we print it

System.out.println(y);

y still points to "old". A new String object was assigned to x which was local to the method func. Therefore, it still prints "old".

score 0 · Answer 4 · answered Oct 20 '13 at 22:44

So, if I declare a string and pass it into a function where I change the value of the string, why doesn't the original string change?

The important thing to know is that you cannot declare a variable whose value is a string -- strings (which are objects) are not values in Java. y is a reference (a pointer to an object). The only types in Java are primitive types and reference types. The type String is a reference type.

Hence, you are passing y, a reference (a pointer to an object). Inside the function, x is likewise a reference (a pointer to an object). Changing the value of a reference (e.g. assigning (=) to x) never has any effect on the object it may have pointed to.

The . operator allows you to access a field of the object pointed to by a reference. Assigning to the result of a . access changes the object pointed to by the reference, not the reference itself. These are very different things.

score -1 · Answer 5 · answered Oct 19 '13 at 06:15

In Java Objects and Strings are treated differently, Strings are kept immutable in JAVA.

To be more clear in your example:

Lets assume, String y is pointing to memory location 2000 and string x is pointing to memory location 2001, so you get the same old value..

In case of objects please refer my example:

http://javaambition.blogspot.kr/2013/08/java-pass-by-value.html

Confused about memory in java

5 Answers5