Default Arguments

Like most of the languages that support imperative programming, a function in PHP may have one or more arguments that have a default value. As a result, such a function may be called without passing any value to it. If there is no value meant to be passed, the function will take its default value for processing. If the function call does provide a value, the default value will be overridden.

functionfun($arg1= val1,$arg2= val2){
   Statements;}

Such a function can be called in different ways −

fun();# Function will use defaults for both argumentsfun($x);# Function passes $x to arg1 and uses default for arg2fun($x,$y);# Both arguments use the values passed

Example 1

Here we define a function called greeting() with two arguments, both having string as their default values. We call it by passing one string, two strings and without any argument.

Open Compiler

<?php  
   function  greeting($arg1="Hello", $arg2="world") {
      echo $arg1 . " ". $arg2 . PHP_EOL;
   }

   greeting();
   greeting("Thank you");
   greeting("Welcome", "back");
   greeting("PHP");
?>

It will produce the following output −

Hello world
Thank you world
Welcome back
PHP world

Example 2

You can define a function with only some of the arguments with default value, and the others to which the value must be passed.

Open Compiler

<?php  
   function  greeting($arg1, $arg2="World") {
      echo $arg1 . " ". $arg2 . PHP_EOL;
   }

   # greeting(); ## This will raise ArgumentCountError
   greeting("Thank you");
   greeting("Welcome", "back");
?>

It will produce the following output −

Thank you World
Welcome back

The first call (without argument) raises ArgumentCountError because you must pass value for the first argument. If only one value is passed, it will be used by the first argument in the list.

However, if you declare arguments with default before arguments without defaults, such function can be only called if values for both are passed. You cannot have a situation where the first argument uses the default, and the second using the passed value.

The greeting() function now has $arg1 with default and $arg2 without any default value.

functiongreeting($arg1="Hello",$arg2){echo$arg1." ".$arg2.PHP_EOL;}

If you pass a string “PHP” −

greeting("PHP");

with the intension to print the result as “Hello PHP”, the following error message will be displayed.

PHP Fatal error: Uncaught ArgumentCountError: Too few arguments to function 
greeting(), 1 passed in hello.php on line 10 and exactly 2 expected

Explore our latest online courses and learn new skills at your own pace. Enroll and become a certified expert to boost your career.

Example 3

Let’s define a function percent() that calculates the percentage of marks in three subjects.

Assuming that the marks in each subject are out of 100, the $total argument in the function definition is given a default value as 300.

Open Compiler

<?php
   function  percent($p, $c, $m, $ttl=300) {
      $per = ($p+$c+$m)*100/$ttl;
      echo "Marks obtained: \n";
      echo "Physics = $p Chemistry = $c Maths = $m \n";
      echo "Percentage = $per \n";
   }
   percent(50, 60, 70);
?>

It will produce the following output −

Marks obtained: 
Physics = 50 Chemistry = 60 Maths = 70 
Percentage = 60

However, if the maximum marks in each subject is 50, then you must pass the fourth value to the function, otherwise the percentage will be calculated out of 300 instead of 150.

Open Compiler

<?php
   function  percent($p, $c, $m, $ttl=300) {
      $per = ($p+$c+$m)*100/$ttl;
      echo "Marks obtained: \n";
      echo "Physics = $p Chemistry = $c Maths = $m \n";
      echo "Percentage = $per \n";
   }
   percent(30, 35, 40, 150);
?>

It will produce the following output −

Marks obtained: 
Physics = 30 Chemistry = 35 Maths = 40 
Percentage = 70