Path: csiph.com!fu-berlin.de!uni-berlin.de!individual.net!not-for-mail
From: "J.O. Aho" <user@example.net>
Newsgroups: comp.lang.php
Subject: Re: value returned instead change the value of the variable
Date: Fri, 16 Oct 2020 20:51:12 +0200
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On 16/10/2020 13.23, alex wrote:
> // DESIGN 1
> echo "It doesn't work (return 1 vs array), but I like 'DESIGN 1' ";
> $arr = [1, 2];
> print_r(
>      array_shift($arr)
> );
> 
 > Output:
 >
 > It doesn't work (return 1 vs array), but I like 'DESIGN 1': 1
 >

If you look at your original array, the cell 0 has an integer, so the 
return type will be int. See https://www.php.net/array_shift it say's:

	Return Values
		Returns the shifted value, or NULL if array is empty or
		is not an array.

If you prompt want an array, either you make it to one afterwards or 
store the values as single cell arrays in the main array

Alt 1:
	array(array_shit($arr));

Alt 2: $arr = [[1],[2]];



> // DESIGN 2
> echo "It works, but I don't like 'DESIGN 2': ";
> $arr = [1, 2];
> array_shift($arr);
> print_r(
>      $arr
> );
> 
> Output:
> 
> It works, but I don't like 'DESIGN 2': Array
> (
>      [0] => 2
> )

Sure we compare apples and oranges, the $arr will be an array even if 
you shift out the last value from it.

try this:

  $arr = [1, 2];
  array_shift($arr);
  print_r($arr);
  array_shift($arr);
  print_r($arr);
  echo "This will confuse you even more\n";
  print_r(array_shift($arr));


-- 
  //Aho