PHP syntax for dereferencing function result

Question

Background

In every other programming language I use on a regular basis, it is simple to operate on the return value of a function without declaring a new variable to hold the function result.

In PHP, however, this does not appear to be so simple:

example1 (function result is an array)

<?php 
function foobar(){
    return preg_split('/\s+/', 'zero one two three four five');
}

// can php say "zero"?

/// print( foobar()[0] ); /// <-- nope
/// print( &foobar()[0] );     /// <-- nope
/// print( &foobar()->[0] );     /// <-- nope
/// print( "${foobar()}[0]" );    /// <-- nope
?>

example2 (function result is an object)

<?php    
function zoobar(){
  // NOTE: casting (object) Array() has other problems in PHP
  // see e.g., http://stackoverflow.com/questions/1869812
  $vout   = (object) Array('0'=>'zero','fname'=>'homer','lname'=>'simpson',);
  return $vout;
}

//  can php say "zero"?       
//  print zoobar()->0;         //  <- nope (parse error)      
//  print zoobar()->{0};       //  <- nope                    
//  print zoobar()->{'0'};     //  <- nope                    
//  $vtemp = zoobar();         //  does using a variable help?
//  print $vtemp->{0};         //  <- nope     

Can anyone suggest how to do this in PHP?

Answer 1

This is too far-fetched, but if you really NEED it to be in one line:

return index0( $foo->getBarArray() );

/* ... */

function index0( $some_array )
{
  return $some_array[0];
}

Answer 2

You could use references:

$ref =& myFunc();
echo $ref['foo'];

That way, you're not really creating a duplicate of the returned array.

Answer 3

You can't chain expressions like that in PHP, so you'll have to save the result of array_test() in a variable.

Try this:

function array_test() {
  return array(0, 1, 2);
}

$array = array_test();
echo $array[0];

Answer 4

If it is just aesthetic, then the Object notation will work if you return an object. As far as memory management goes, no temporary copy if made, only a change in reference.

Answer 5

As others have mentioned, this isn't possible. PHP's syntax doesn't allow it. However, I do have one suggestion that attacks the problem from the other direction.

If you're in control of the getBarArray method and have access to the PHP Standard Library (installed on many PHP 5.2.X hosts and installed by default with PHP 5.3) you should consider returning an ArrayObject instead of a native PHP array/collection. ArrayObjects have an offetGet method, which can be used to retrieve any index, so your code might look something like

<?php
class Example {
    function getBarArray() {
        $array = new ArrayObject();
        $array[] = 'uno';
        $array->append('dos');
        $array->append('tres');
        return $array;
    }
}

$foo = new Example();
$value = $foo->getBarArray()->offsetGet(2);

And if you ever need a native array/collection, you can always cast the results.

//if you need 
$array = (array) $foo->getBarArray();

Answer 6

PHP can not access array results from a function. Some people call this an issue, some just accept this as how the language is designed. So PHP makes you create unessential variables just to extract the data you need.

So you need to do.

$var = foobar();
print($var[0]);