Passing a String by Reference in Java

Question

I am used to doing the following in C   void main         String zText           fillString zText       printf zText      void fillString String zText        zText     foo       And the output is   foo   However  in Java  this does not seem to work  I assume because the String object is copied instead of passed by referenced  I thought Strings were objects  which are always passed by reference    What is going on here

User · Answer

What is happening is that the reference is passed by value, i.e., a copy of the reference is passed. Nothing in java is passed by reference, and since a string is immutable, that assignment creates a new string object that the copy of the reference now points to. The original reference still points to the empty string.

This would be the same for any object, i.e., setting it to a new value in a method. The example below just makes what is going on more obvious, but concatenating a string is really the same thing.

void foo( object o )
{
    o = new Object( );  // original reference still points to old value on the heap
}

User · Answer

You have three options    Use a StringBuilder   StringBuilder zText   new StringBuilder     void fillString StringBuilder zText    zText append   foo       Create a container class and pass an instance of the container to your method   public class Container   public String data    void fillString Container c    c data     foo      Create an array   new String   zText   new String 1   zText 0         void fillString String   zText    zText 0      foo        From a performance point of view  the StringBuilder is usually the best option

User · Answer

Strings are immutable in Java

User · Answer

String is immutable in java  you cannot modify change  an existing string literal object    String s  Hello   s s  hi    Here the previous  reference s is replaced by the new refernce s pointing to value  HelloHi    However  for bringing mutability we have StringBuilder and StringBuffer    StringBuilder s new StringBuilder    s append  Hi     this appends the new value  Hi  to the same refernce s

User · Answer

In Java nothing is passed by reference  Everything is passed by value  Object references are passed by value  Additionally Strings are immutable  So when you append to the passed String you just get a new String  You could use a return value  or pass a StringBuffer instead

User · Answer

For someone who are more curious   class Testt       static void Display String s   String varname           System out println varname     variable data      s        address hashmap        s hashCode                static void changeto String s   String t           System out println  entered function            Display s    s            s   t           Display s  s            System out println  exiting function               public static void main String args             String s     hi            Display s  s            changeto s  bye            Display s  s                 Now by running this above code you can see how address hashcodes change with String variable s      a new object is allocated to variable s in function changeto when s is changed

User · Answer

All arguments in Java are passed by value  When you pass a String to a function  the value that s passed is a reference to a String object  but you can t modify that reference  and the underlying String object is immutable   The assignment  zText    foo    is equivalent to   zText   new String zText    foo      That is  it  locally  reassigns the parameter zText as a new reference  which points to a new memory location  in which is a new String that contains the original contents of zText with  foo  appended   The original object is not modified  and the main   method s local variable zText still points to the original  empty  string   class StringFiller     static void fillString String zText        zText     foo       System out println  Local value      zText          public static void main String   args        String zText           System out println  Original value      zText       fillString zText       System out println  Final value      zText           prints   Original value  Local value  foo Final value    If you want to modify the string  you can as noted use StringBuilder or else some container  an array or an AtomicReference or a custom container class  that gives you an additional level of pointer indirection  Alternatively  just return the new value and assign it   class StringFiller2     static String fillString String zText        zText     foo       System out println  Local value      zText       return zText         public static void main String   args        String zText           System out println  Original value      zText       zText   fillString zText       System out println  Final value      zText           prints   Original value  Local value  foo Final value  foo   This is probably the most Java-like solution in the general case -- see the Effective Java item  Favor immutability    As noted  though  StringBuilder will often give you better performance -- if you have a lot of appending to do  particularly inside a loop  use StringBuilder   But try to pass around immutable Strings rather than mutable StringBuilders if you can -- your code will be easier to read and more maintainable  Consider making your parameters final  and configuring your IDE to warn you when you reassign a method parameter to a new value

User · Answer

String is an immutable object in Java  You can use the StringBuilder class to do the job you re trying to accomplish  as follows   public static void main String   args        StringBuilder sb   new StringBuilder  hello  world         System out println sb       foo sb       System out println sb       public static void foo StringBuilder str        str delete 0  str length         str append  String has been modified        Another option is to create a class with a String as a scope variable  highly discouraged  as follows   class MyString       public String value     public static void main String   args        MyString ms   new MyString        ms value    Hello  World        public static void foo MyString str        str value    String has been modified

User · Answer

The answer is simple  In java strings are immutable  Hence its like using  final  modifier  or  const  in C C     So  once assigned  you cannot change it like the way you did   You can change which value to which a string points  but you can NOT change the actual value to which this string is currently pointing   Ie  String s1    hey   You can make s1    woah   and that s totally ok  but you can t actually change the underlying value of the string  in this case   hey   to be something else once its assigned using plusEquals  etc   ie  s1      whatup     hey whatup     To do that  use the StringBuilder or StringBuffer classes or other mutable containers  then just call  toString   to convert the object back to a string    note  Strings are often used as hash keys hence that s part of the reason why they are immutable

User · Answer

objects are passed by reference  primitives are passed by value   String is not a primitive  it is an object  and it is a special case of object    This is for memory-saving purpose  In JVM  there is a string pool  For every string created  JVM will try to see if the same string exist in the string pool  and point to it if there were already one   public class TestString       private static String a    hello world       private static String b    hello world       private static String c    hello      world       private static String d   new String  hello world         private static Object o1   new Object        private static Object o2   new Object         public static void main String   args                System out println  a  b    a    b            System out println  a  c    a    c            System out println  a  d    a    d            System out println  a equals d     a equals d             System out println  o1  o2    o1    o2             passString a           passString d              public static void passString String s                System out println  passString    a    s                  OUTPUT     a  b true a  c true a  d false a equals d  true o1  o2 false passString true passString false   the    is checking for memory address  reference   and the  equals is checking for contents  value

User · Answer

This works use StringBuffer  public class test    public static void main String   args     StringBuffer zText   new StringBuffer          fillString zText       System out println zText toString          static void fillString StringBuffer zText        zText  append  foo          Even better use StringBuilder  public class test    public static void main String   args     StringBuilder zText   new StringBuilder          fillString zText       System out println zText toString          static void fillString StringBuilder zText        zText  append  foo

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String is a special class in Java  It is Thread Safe which means  Once a String instance is created  the content of the String instance will never changed     Here is what is going on for   zText     foo     First  Java compiler will get the value of zText String instance  then create a new String instance whose value is zText appending  foo   So you know why the instance that zText point to does not changed  It is totally a new instance  In fact  even String  foo  is a new String instance  So  for this statement  Java will create two String instance  one is  foo   another is the value of zText append  foo    The rule is simple  The value of String instance will never be changed   For method fillString  you can use a StringBuffer as parameter  or you can change it like this   String fillString String zText        return zText     foo

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For passing an object  including String  by reference in java  you might pass it as member of a surrounding adapter  A solution with a generic is here   import java io Serializable   public class ByRef lt T extends Object gt  implements Serializable       private static final long serialVersionUID   6310102145974374589L       T v       public ByRef T v                this v   v             public ByRef                 v   null             public void set T nv                v   nv             public T get                 return v            ------------------------------------------------------------------      static void fillString ByRef lt String gt  zText                zText set zText get      foo               public static void main String args                  final ByRef lt String gt  zText   new ByRef lt String gt  new String               fillString zText           System out println zText get

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java lang String is immutable   I hate pasting URLs but https   docs oracle com javase 10 docs api java lang String html is essential for you to read and understand if you re in java-land

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Aaron Digulla has the best answer so far  A variation of his second option is to use the wrapper or container class MutableObject of the commons lang library version 3    void fillString MutableObject lt String gt  c    c setValue c getValue      foo        you save the declaration of the container class  The drawback is a dependency to the commons lang lib  But the lib has quite a lot of useful function and almost any larger project i have worked on used it

[java] Passing a String by Reference in Java?

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