Function Parameters Are Passed as Local Copies

JavaScript Tutorials - Herong's Tutorial Examples

∟Function Parameters Are Passed as Local Copies - Example

This section provides a tutorial example on swapping original primitive and object variables in a function.

To prove that JavaScript function always create local copies for object parameters, I wrote the following tutorial example to try to swap two objects:

<html>
<!-- Swap_Function.html
   Copyright (c) 2008 HerongYang.com. All Rights Reserved.
-->
<head><title>Swap Function</title></head>
<body>
<pre>
<script type="text/javascript">
// Defining a swap function
function swap(left, right) {
   var temp = left;
   left = right;
   right = temp;
   document.write("   Inside swap(): "+left+" | "+right+"\n");
   document.write("   Type of argument: "+(typeof left)+"\n"); 
}

   var first = "Dog";
   var second = "Cat";
   document.write("\nTest 1: Swapping two string variables\n");
   document.write("   Before call: "+first+" | "+second+"\n");
   swap(first,second);
   document.write("   After call: "+first+" | "+second+"\n");

   var one = new String("White");
   var two = new String("Black");
   document.write("\nTest 2: Swapping two string objects\n");
   document.write("   Before call: "+one+" | "+two+"\n");
   swap(one,two);
   document.write("   After call: "+one+" | "+two+"\n");
</script>
</pre>
</body>
</html>

The output shows that function parameters are always passed as local copies of original variables:

Test 1: Swapping two string variables
   Before call: Dog | Cat
   Inside swap(): Cat | Dog
   Type of argument: string
   After call: Dog | Cat

Test 2: Swapping two string objects
   Before call: White | Black
   Inside swap(): Black | White
   Type of argument: object
   After call: White | Black