[PHP-DEV] PHP 4.0 Bug #7578 Updated: next() and current() do not return referenceing arrays From: waldschrott <email protected>
Date: 11/01/00

ID: 7578
Updated by: waldschrott
Reported By: mog <email protected>
Status: Closed
Bug Type: Arrays related
Assigned To:
Comments:

you simply can´t do it that way, please read the manual on foreach() etc. and that additional "&" in
> $array = array(&$array2);
has absolutely no effect here, at least not what you expect it to do (please also read "references explained")

> print next(current($array))."<br>"; // returns 1, correct > but the internal
> pointer is only moved in the copy current() returned

sure, current($array) returns a copy and thus all results you´ve mentioned are fine

Previous Comments:
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[2000-11-01 19:29:27] mog <email protected>
i hope you can see what is wrong in the code below

<?PHP
$array2 = array(0,1,2);
$array = array(&$array2);

print current(current($array))."<br>"; // returns 0

print "<b>Try 1!</b><br>";
print next(current($array))."<br>"; // returns 1, correct but the internal pointer is only moved in the copy current() returned
print current(current($array))."<br>"; // returns 0, wrong, should be 1
print current($array2)."<br>"; //returns 0, wrong, should be 1

print "<b>Try 2!</b><br>";
print next($array[0])."<br>";
print current(current($array))."<br>"; // returns 0, wrong, should be 1
print current($array[0])."<br>"; // returns 1, yes, correct! but the code above still didn't work!
?>

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Full Bug description available at: http://bugs.php.net/?id=7578 

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