php store array in array by reference

Question

I have the following problem. There are some php lines.

In the var_dump of $testArrayA, the "def" entry with test2 is NOT there because it was added after $testArrayB was added to $testArrayA.

It seems to me that in my case, that $testArrayB is not stored by reference in $testArrayA. How can I store it per reference, what do I have to make to have "def" entry in the var_dump?

Thanks alot in advance

$testArrayA = [];
$testArrayB = [];
$testArrayB["ghi"] = "test1";
$testArrayA["abc"] = $testArrayB;
$testArrayB["def"] = "test2";

The var_dump:

array(1) {
  ["abc"]=>
    array(1) {
       ["ghi"]=>
       string(5) "test1"
      }
}

shawndreck · Accepted Answer · 2013-01-23 04:25:34Z

7

This is simply a matter of passing by reference:

$testArrayA = [];
$testArrayB = [];
$testArrayB["ghi"] = "test1";
$testArrayA["abc"] = &$testArrayB;
$testArrayB["def"] = "test2";

var_dump($testArrayA);

array (size=1)
'abc' => &
array (size=2)
  'ghi' => string 'test1' (length=5)
  'def' => string 'test2' (length=5)

answered Jan 23, 2013 at 4:25

shawndreck

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xdazz · Accepted Answer · 2013-01-23 04:25:33Z

6

Use:

$testArrayA["abc"] = &$testArrayB;

Note:

Different with C's pointer, references in PHP are a means to access the same variable content by different names.

answered Jan 23, 2013 at 4:25

xdazz

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user1283182 · Accepted Answer · 2013-01-23 04:26:25Z

1

in the php manual

Array assignment always involves value copying. Use the reference operator to copy an array by reference.

answered Jan 23, 2013 at 4:26

user1283182

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Jhonny · Accepted Answer · 2013-01-23 04:36:50Z

0

$testArrayA = null;
$testArrayB = null;
$testArrayB["ghi"] = "test1";
$testArrayA["abc"] = $testArrayB;
$testArrayB["def"] = "test2";
print_r($testArrayA);
echo ("<br>");
print_r($testArrayB);


Array ( [abc] => Array ( [ghi] => test1 ) ) 
Array ( [ghi] => test1 [def] => test2 )

The 'def' entry is a different reference with the 'ghi', but they all belong testArrayB.

$testArrayA["abc"] = $testArrayB;

This code only is value reference ,not address reference.

answered Jan 23, 2013 at 4:36

Jhonny

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