Passing a string to a function

Question

I would like to ask a really simple question. I am just passing a String object to a function but the result is weird. I supposed that because I am passing an object(by reference) the result should have be " Here" and not "Hello". Why is this happening?

public class MainStr
{

/**
 * @param args
 */
public static void main(String[] args)
{
    String str = "Hello!";
    System.out.println(str);
    changeString(str);
    System.out.println(str);


}

static void changeString(String str)
{
    str=" HERE";
}

}

Thanks.

EDITED:

public class MainStr
{

String str;
/**
 * @param args
 */
public static void main(String[] args)
{
    MainStr ms = new MainStr();
    ms.str = "Hello!";
    System.out.println(ms.str);
    changeString(ms);
    System.out.println(ms.str);


}

static void changeString(MainStr ms)
{
    ms.str=" HERE";

}

}

If that is the case then why if I pass it inside a wrapper object is working? The wrapper object it is not by reference?

PS: Why this is working that way? String is an object. Why do I need another wrapper Object to change an OBJECT! What if I want to pass something by reference? Is that possible?

Answer 1

Java passes object references by value this means when you call a function the following happens.

 String origStr = "Some String"

 // origStr -> "Some String "

When the object is passed to the value it looks like this

 //origStr -> "Some String " <- passedStr

Then in the function you reset what passedStr points to

 //origStr -> "Some String "
 //passedStr -> "Other String"

The original string still refers to "Some String" but the copy of it now refers to "Other String"

When you return from the function the original string reference still points at "Some String"

// EDIT: Why can you modify objects referred to by another object?

Lets say I have StringWrapper, which is a class which contains a string.

To java that looks like this

origStr -> strWrpA -> "Some String"

When you pass it to a function the following happens just like above

origStrWrp -> strWrpA -> "Some String"
passedStrWrp --/^

Notice how both variables point at strWrpA in memory. Now if i decided to modify the strWrpA.myString Java will follow the pointer to the actual string in the wrapper. The pointer to strWrpA will remain unchanged. This means I can change the contents of the object even though I'm accessing it through a copy of the original reference.

                        "Some String "
origStrWrp -> strWrpA -> "I got Changed!"
passedStrWrp --/^

You would have the same problem as you have without a wrapper if you did this

//Does Nothing
function (origStrWrp)
{
origStrWrp=new StringWrapper("Other String") // This only reassigns the copy of the reference and would give you the same problem
}
//Changes internal object
function (origStrWrp)
{
origStrWrp.myString = "Other String" // This doesn't change the reference, it changes the object that it refers to.
}

Answer 2

In Java references are passed by value. Here is popular SO discussion to understand this topic. The reference str is scoped to changeString method only because of passing reference as value.

Answer 3

Strings are immutable in Java. In order for this to work you need to return the newly created string instance from the hangeString method.

static String changeString(String str)
{
    return " HERE";
}

Answer 4

Doing str=" HERE"; actually changes the object that str points to. It doesn't change the original object. If you want to modify the original, put it in a wrapper object.

Example:

public class MainStr {

    static class StringWrapper {
        public String str;
    }

    public static void main(String[] args) {
        StringWrapper foo = new StringWrapper();
        foo.str = "Hello!";
        System.out.println(foo.str);
        changeString(foo);
        System.out.println(foo.str);
    }

    static void changeString(StringWrapper str)
    {
        str.str = "Bye!";
    }
}

Answer 5

Yes, Java is passing class arguments by reference. Impossible to tell it to do other way.

You can wrap your string into another class or use StringBuilder from the library.

P.S.

str=" HERE";

just replaces old reference to a new one pointing to " HERE". This does not touch old reference str outside a function.

P.P.S.

Probably my terminology is wrong. So, in plain words, Java variable of type String contains a reference to that string. This reference is passed into function. Assignment is changing a reference, not original object.

Answer 6

you are mixing things. although both variables are called 'str' they are in different 'scopes'

the first is only valid inside your main method while the second is only valid inside your changeString method.

it seems that if you want to accomplish your goal of using the changeString to make changes available within the main method; you will have to:

a) create a variable outside the methods - so that it will have a class (global) scope:

favorite share [g+] share [fb] share [tw]

I would like to ask a really simple question. I am just passing a String object to a function but the result is weird. I supposed that because I am passing an object(by reference) the result should have be " Here" and not "Hello". Why is this happening?

public class MainStr
{
String str = null;

public static void main(String[] args)
{
    str = "Hello!";
    System.out.println(str);
    changeString();  // there is no argument to supply
    System.out.println(str);


}

static void changeString()
{
    str= str + " HERE"; // variable is changed here
}

}

b) enable changeString to return a string, and use that:

public class MainStr 
{

/**
 * @param args
 */
public static void main(String[] args)
{
    String str = "Hello!";
    System.out.println(str);
    str = changeString(str);   // str has return value assigned to it!
    System.out.println(str);   // otherwise it would print out 'Hello!'


}

static String changeString(String str)
{
    str = str.concat(" HERE");
    return str;

}
}

NOTE: in the second option (b) you still two variable with different scopes and same name.

while in the first option (a) there is only one variable in the whole

hope this clarifies your doubts.

good luck!

Nicolas