Unexpected result for Var Dump for $products

The removed variable is just a placeholder, somewhere to assign the method to, and that''s it's only function. It's like a paperclip holding a piece of paper together.

The var_dump isn't interested in the removed variable and all it does is display the contents of the variable that's passed in, which in this case is $products. :-)

HI Jonathan. Thanks for the reply.

Yes, I agree that's what it should do. But if you comment out the line that sets up the removed variable, you will see a different result for the var dump of the products variable.

So, test it like this: var_dump($products);

and then like this: $removed = array_shift($products); var_dump($products);

You will get a different result for the var dump. In the first instance it shows the array with the associate key that we assigned to it in the array. In the 2nd instance it appears to remove the first product and reset the array keys.

Hi Jonathan. Sorry, I must not be explaining this very well. Just to clarify: I understand what the array_shift() does. What I'm not getting here is that we're locking the shift array up in a new variable called $removed, but then we're going back to the same var dump as before i.e. var_dump($products)

I would have thought that to access the result of the array shift function you would need to var dump the variable that has created the array_shift and not the original array that doesn't have the array_shift applied.

So, to see the result of the array_shift function I would have thought we'd have to do one of these two:

This: var_dump(array_shift($products));

or this: $removed = array_shift($products); var_dump($removed);

Do you see what I'm getting at here?

In the example you've given above, yes it makes sense that it would show a shifted array. I understand that it works from the inside out, but that's not my point. Maybe this example will clarify what I mean. See the difference between these two:

<?php 
   $removed = array_shift($products);
     var_dump($products);
?>

<?php 
   array_shift($products);
     var_dump($products);
?>

In the 2nd example, I didn't put the array shift into a variable, so I would expect the dump to show the shifted array. In the first example, I've locked the shifted array up into a variable. I'm still not getting how we can access the shifted array without doing a var dump of the $removed variable.

Yes, I can see that what he's done does produce the result he's looking for, but I can't understand the logic.

Thanks a lot, Jonathan.