Does the PHP current() function return a copy or reference of array?

Question

I created a simple test case that replicates an issue I'm having.

I'm walking through a two dimensional array using the next() and current() functions, and am wanting to setup the array pointer to a specific location. So, given a 2D array with a variable name of $food with the following array structure:

array
  0 => <-- POINTER LOCATION
    array
      0 => string 'apple' <-- POINTER LOCATION
      1 => string 'orange'
  1 => 
    array
      0 => string 'onion'
      1 => string 'carrot'

... And the following code snippet:

// move the inner array's pointer once
$should_be_orange = next(current($food));

// now check that inner array's value
$should_still_be_orange = current(current($food));

... Why is the value of $should_be_orange "orange" but the value of $should_still_be_orange "apple"? Is this because the current() function returns a copy of the inner array, who's pointer gets iterated, and then destroyed (leaving the original array untouched)? Or am I simply doing something wrong that I'm not catching?

At the root of the question, how do you move the inner array's pointer given that you don't know the outer array's key (and must use the current() function to obtain the outer array's pointer location)?

Maxim Krizhanovsky · Accepted Answer · 2011-11-15 21:36:43Z

2

Actually current() returns an element from the array. In your case, this element is also an array, and that's why next() is working at all in your code. Your next() do not work on the $food array, but on copy of $food[0], returned by current()

answered Nov 15, 2011 at 21:36

Maxim Krizhanovsky

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1 Comment

Jared Cobb Over a year ago

Okay, exactly what I suspected. Assuming I always have a two dimensional array (like the example), do you have a suggestion on how I can move the pointer of the inner array?

BenMorel · Accepted Answer · 2014-01-06 12:04:12Z

1

You can't pass function in arguments, you can must only variables, because argument is reference:

function current(&$array) {...}
function next(&$array) {...}

so proper syntax is:

// move the inner array's pointer once
$tmp = current($food);
$should_be_orange = next($tmp);

// now check that inner array's value
$tmp = current($food);
$should_still_be_orange = current($tmp);
                 ^^^^^^ NO! It should be "apple" ! When you do next($tmp) it will be orange !

Demo: http://codepad.viper-7.com/YZfEAw

Documentation:

When you are learning PHP you should display ALL errors using command:

error_reporting(E_ALL);

Using this u should receive notice:

Strict Standards: Only variables should be passed by reference in (...) on line (...)

(i think this answer need review cause of english gramma)

edited Jan 6, 2014 at 12:04

BenMorel

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answered Nov 15, 2011 at 21:33

Peter

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4 Comments

mario Over a year ago

The documentation is not in sync with the current implementation. You are supposed to pass variables to next() and current(), but expressions / array constants work too.

Peter Over a year ago

Yes it will work, but it will create Strict standards error as is mention above. Passing function instead of variable is main problem here.

mario Over a year ago

In which version did you test that? Because it does neither in PHP 5.3 nor 5.4 for me. -- And no, the main problem is that no reference is returned, as OP already suspected.

Peter Over a year ago

Yes! All info is in documentation.

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