PHP define a variable if it's not passed to a function

Hey there!

I've faced a similar situation before, and there's a handy way to accomplish this in PHP. What you can do is use the `func_num_args()` function combined with an `if` statement to check if the `$optionalVar` parameter was passed or not.

Here's an example of how you can implement it within your `myFunction()`:

php
function myFunction($optionalVar = null) {

    // Check if $optionalVar was passed as an argument

    if (!func_num_args()) {

        $newVar = "Default value";

    }



    // Rest of your code goes here

}

Hey everyone,

I've encountered a similar situation in my PHP projects, and I found an alternative approach that might be useful to you. Instead of relying on `func_num_args()`, you can utilize the `func_get_args()` function.

Here's an example of how you can handle the scenario:

php
function myFunction($optionalVar = null) {

    // Use func_get_args() to fetch all passed arguments

    $args = func_get_args();



    // Check if $optionalVar was explicitly passed

    if (count($args) === 1) {

        $newVar = "Default value";

    }



    // Rest of your code goes here

}

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