String Doubt

[Nischal Tanna]

  
 
import java.util.Hashtable;
import java.util.Enumeration;
import java.util.ArrayList;
 
public class Test
{
static String name = "";
public  static void swap(String name)
{
name = "Test1";
System.out.println("Name is "+name);
}
public static void main(String args[])
{
name = "Test2";
System.out.println("Name before = "+name);
swap(name);
System.out.println("Name after = "+name);
 

}
 
}
 

The above code prints Name after = Test2. I thought it will be printing Test1 as String is an Object and hence passed by reference.

[Lionel Badiou]

Hello Nischal,

you may find useful to read this

Hope that helps,

[Amol Fuke]

The name in the method has scope of local so it will as well take name in the method.

[Nischal Tanna]

Originally posted by Amol Fuke:
The name in the method has scope of local so it will as well take name in the method.

Not justified to say that. If thats the case then there will be no such concept of Pass By Reference

[David Harkness]

Originally posted by Nischal Tanna:
there will be no such concept of Pass By Reference

I hate to give anyone bad news, but it's true: Java has no concept of pass-by-reference. Everything is pass-by-value -- even when you're passing a reference.

The name reference has a value which you can think of conceptually as the address of the "Test2" String. When you execute swap(name), a copy of that value is placed onto the stack for the swap() method to access. When swap() changes that value, it's changing the copy of the reference, so it has no effect on the main() method's name reference.

It seems like Java has pass-by-reference because you can alter objects through references passed to methods. Yet this is no different than passing a pointer to a struct to a C function. In that function you can alter the struct's fields but you cannot alter the pointer itself. To do that, you'd have to pass a pointer to the pointer in C or change the argument to a pointer reference in C++.

Neither of these are possible in Java.

[Ryan McGuire]

Originally posted by David Harkness:
I hate to give anyone bad news, but it's true: Java has no concept of pass-by-reference. Everything is pass-by-value -- even when you're passing a reference.

It seems like Java has pass-by-reference because you can alter objects through references passed to methods. Yet this is no different than passing a pointer to a struct to a C function. In that function you can alter the struct's fields but you cannot alter the pointer itself. To do that, you'd have to pass a pointer to the pointer in C or change the argument to a pointer reference in C++.

Semantics question:
But isn't passing a reference by value the exact same thing as passing and object by reference? Isn't that how pass-by-reference is defined?

Neither of these are possible in Java.

Depends on what hoops you're willing to jump through. You could wrap a reference to an object in another object and pass that reference. Let's see are there any good single-object wrappers already? An ArrayList or other Collection seems like overkill for just one object.

Ryan

[James Carman]

No, passing an object by reference and passing a reference by value are not the same thing...

  
public static void someMethod( String param )
{
  param = "Some other string";
}
 
public static void main( String[] args )
{
  String s = "Hello, World!";
  someMethod( s );
  System.out.println( s );
}

If Java were pass by reference, this program would print "Some other string" as opposed to "Hello, World!". Pass by reference would allow you to change which object the calling code is referring to.

[David Harkness]

Originally posted by me:
To do that, you'd have to pass a pointer to the pointer in C or change the argument to a pointer reference in C++.

Neither of these are possible in Java.

Originally posted by Ryan McGuire:
You could wrap a reference to an object in another object and pass that reference.

Yes, but here you've fallen back on pass-by-value: passing the value of a reference to a struct. This is not one of the two pass-by-reference methods above. Yes, it seems like semantics, but at the heart it really isn't. Java has no pass-by-reference.

As an example, pretend that Java didn't have the ++ and -- operators. You could say that Java still has an increment operator because you can use "x = x + 1" and "x += 1", but those really aren't increment operators.

[Adam Vinueza]

I think that in the swap() function the copy of name is directed to point at whatever the String constant expression "Test1" points to. When control leaves the function, the original variable's value is unchanged. It's just like a C function, when you pass a pointer into a function, then when inside the function make it point to something else (instead of dereferencing it and changing its contents). Keep in mind that there's no such thing as dereferencing in Java.