Question

I understand I can define a function like this:

<?php 
    function say_hello() {
        echo "hello world!";
    }

    say_hello();
?>

…and then reuse this function elsewhere as many times as I need by just calling the function by name.

What I don’t understand is “passing data to arguments within the function”. I’m confused what’s meant by within.

From one of the lessons I’ve been studying, I have this function:

<?php 
   function say_helloTWO( $word ) {
       echo " {$word} hello world a second time ";
   }

   say_helloTWO( "my name is mike");
?>

This prints on screen

my name is mike hello world a second time

When I test the function with the argument "my name is mike", it prints to screen, but I don’t understand how this works. The variable $word wasn’t declared anywhere, so, if I take out the "$word" from the echo, then only "hello world a second time” shows without the my name is mike.

This leads me to believe that somewhere within this block of code, $word was defined, is that right?

Answer 1

In your second function $word is being declared in the function definition function say_helloTWO( $word ).

So the function is expecting 1 parameter, which will be assigned to the variable $word for use within that function.

So when you call say_helloTWO( "my name is mike"); you are passing 1 parameter (“my name is mike”) to the function say_helloTWO. This 1 parameter gets assigned to the 1st variable in the function definition, and is therefore available within the function as $word.

Make sense?