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PHP PHP Functions Introducing Functions PHP Function Arguments

Lolo Handricks
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Lolo Handricks
Front End Web Development Techdegree Student 10,706 Points
Posted by Lolo Handricks
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Lolo Handricks
Front End Web Development Techdegree Student 10,706 Points

$arr, $name, $names

I understand how $arr and $name work together, but how does the function know that the $names variable goes together with the $arr and $name variables when it calls the function using the $names variable in the ()?

1 Answer

Chris Shaw
Chris Shaw
26,676 Points
Chris Shaw
Chris Shaw
26,676 Points

Hi Eloise,

As Hampton mentioned this is called passing by value, essentially what this means is we're passing a copy of the array $names to our hello function which when executed becomes $arr as it's now a scoped parameter instead of a global variable. Now that the function has a copy of the array we iterate over it using the foreach function as shown in the video.

Hope that helps.

adrien kovanic
adrien kovanic
5,727 Points
adrien kovanic
adrien kovanic
5,727 Points

hi, I had the same question, to go a little deeper, I don't understand why it is selecting this array 'names' how come that if I add on other array under let's say last_name it will not call it in the function?

thanks a lot

Anna Donlin
Anna Donlin
687 Points
Anna Donlin
Anna Donlin
687 Points

Hello, In layman's terms: the code on line 4 of the lesson if(is_array($arr)){ applies the name of the array $names to $arr. I did the lesson, renaming the $names array to $gypsies just to clarify it to myself. It worked. The foreach($arr as $name) statement on line 5 applies each piece of the array to the $name variable. It's circular, since the array is defined after the function. It's not intuitive to then associate the array back to the is_array line in the function.

Hope this helps!