Pass-by-value Semantics in Java Applications

By Peter Haggar
2003-11-24

Writing a swap method

Given what we know about how parameters are passed, writing a swap function in C++ can be accomplished in different ways. A swap function using pointers, which are passed by value, looks like this:Listing 5: Swap function using pointers



#include 

#include 



void swap(int *a, int *b);



int main (int argc, char** argv)

{

  int val1, val2;

  val1 = 10;

  val2 = 50;

  swap(&val1, &val2);

  return 0;

}



void swap(int *a, int *b)

{

  int temp = *b;

  *b = *a;

  *a = temp;

}

A swap function using references, which are passed by reference, looks like this:

Listing 6: Swap function using references



#include 

#include 



void swap(int &a, int &b);



int main (int argc, char** argv)

{

  int val1, val2;

  val1 = 10;

  val2 = 50;

  swap(val1, val2);

  return 0;

}



void swap(int &a, int &b)

{

  int temp = b;

  b = a;

  a = temp;

}

Both C++ code examples swap the values as expected. If Java applications used pass by reference, the following swap method would work like the C++ examples:

Listing 7: Java swap function if pass by reference worked like in C++



class Swap

{

  public static void main(String args[])

  {

    Integer a, b;



    a = new Integer(10);

    b = new Integer(50);



    System.out.println("before swap...");

    System.out.println("a is " + a);

    System.out.println("b is " + b);

    swap(a, b);

    System.out.println("after swap...");

    System.out.println("a is " + a);

    System.out.println("b is " + b);

  }



  public static void swap(Integer a, Integer b)

  {

    Integer temp = a;

    a = b;

    b = temp;

  }

}

Because the Java application passes all parameters by value, this code does not work and produces the following output:

Listing 8: Output from Listing 7



before swap...

a is 10

b is 50

after swap...

a is 10

b is 50

So how do you write a method in a Java application to swap the values of two primitive types or two object references? Because a Java application passes all parameters by value, you cannot. To swap the values, you must do so inline, outside of a method call.

Tutorial Pages:
» Pass-by-value semantics in Java applications
» The key point
» Parameter passing in C++ and Java applications
» Examples
» Writing a swap method
» Resources

First published by IBM DeveloperWorks

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