Using Pass-by-Value Arguments for References

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This section provides a tutorial example on how to references on pass-by-value arguments. Using a reference expression as a pass-by-value argument allows you to modify the original variable in the calling code.

In the previous section, we tested on passing regular variables to functions using the pass-by-value mode. But what will happen if we pass references to functions using the pass-by-value mode? To find out the answer, I wrote this tutorial example script:

<?php # PassByValueWithReference.php
# Copyright (c) 2003 by Dr. Herong Yang. http://www.herongyang.com/
# 
function swap($left, $right) {
   $temp = $left;
   $left = $right; 
   $right = $temp;
   print("    Swapped in function: ".getString($left, $right)."\n");
}

   print("\n 1. Passing two variables:\n");
   $x = "Dog"; $y = "Cat";
   print("    Before call: ". getString($x, $y) ."\n");
   swap($x, $y);
   print("    After call: ". getString($x, $y) ."\n");

   print("\n 2. Passing two references:\n");
   $y = "Bob"; $y = "Tom";
   print("    Before call: ". getString($x, $y) ."\n");
   swap(&$x, &$y);
   print("    After call: ". getString($x, $y) ."\n");

   print("\n 3. Passing two references in variables:\n");
   $a = "One"; $b = "Two";
   $x = &$a; $y = &$b;
   print("    Before call: ". getString($a, $b) ."\n");
   print("    Before call: ". getString($x, $y) ."\n");
   swap($x, $y);
   print("    After call: ". getString($a, $b) ."\n");
   print("    After call: ". getString($x, $y) ."\n");

function getString($left, $right) {
   if (is_scalar($left)) {
      return "$left | $right";
   } else {
      return NULL;
   }
}
?>

If you run this tutorial example, you should get:

 1. Passing two variables:
    Before call: Dog | Cat
    Swapped in function: Cat | Dog
    After call: Dog | Cat

 2. Passing two references:
    Before call: Dog | Tom
    Swapped in function: Tom | Dog
    After call: Tom | Dog

 3. Passing two references in variables:
    Before call: One | Two
    Before call: One | Two
    Swapped in function: Two | One
    After call: One | Two
    After call: One | Two

Do you see anything interesting in the output?

There is nothing strange in test 1. Pass-by-value on regular variables left no changes on original variables.

But in test 2, pass-by-value on references caused original variables to be swapped. I think I know why. The swap() function received copies of references, &$x and &$y, swapping references actually swapped the original variables, $x and $y.

But in test 3, pass-by-value on references stored in variables caused original variables un-touched. Initially, I thought test 3 should behave like test 2. But by following the pass-by-value logic, I think I know why it behaves differently. In the call, swap($x, $y), copies of values of $x and $y are passed. Since $x and $y are reference variables, values of $x and $y are values of $a and $b, "One" and "Two". So copies of "One" and "Two" are passed into the function. Of course, swapping these copies will have no impact on original values of $a and $b (or $x and $y, since they are aliases of $a and $b).

Conclusion, If you call a function with a reference expression as a pass-by-value argument, you are equivalently using this argument as a pass-by-reference argument. Changes on this argument inside the function will be applied to the original variable in the calling code.

Last update: 2005.

Sections in This Chapter

What Is a Function?

"function" Statements - Defining Functions

Function Call Operations

Passing Arguments to Functions

Example of Passing Arguments by Values

Using Pass-by-Value Arguments for References

Example of Passing Arguments by References

Variable-Length Argument Lists

Providing Default Values to Argument Variables

Returning Values from Functions

Returning References from Functions