Trimming string variables by reference in PHP5

Question

I saw another post suggesting using this statement to trim string variables contained in the array:

$_POST=array_map('trim', $_POST);

However, if in the first place, the strings are not contained in an array, I would like to have a trim function that can be used like this:

$a='  aaa ';
$b='  bbb ';
$c='  ccc ';
trimAll($a,$b,$c); //arbitrary number of string variables can be passed

I tried to write a function like this:

function trimAll() {

    $args = &func_get_args();
    foreach($args as &$arg) {
        if(isset($arg) && is_string($arg))
            $arg=&trim($arg);
    }
      //no return value is required
}

But without success, the string variables do not get trimmed after function return.

Why and how can this be done??

possible duplicate of [PHP: variable-length argument list by reference?](http://stackoverflow.com/questions/1925253/php-variable-length-argument-list-by-reference) — newacct, Oct 07 '11 at 09:20

score 4 · Accepted Answer · answered Oct 29 '09 at 01:15

you cannot pass variable number of parameters by reference. As a workaround, try something like

list($a, $b, $c) = array_map('trim', array($a, $b, $c));

better yet, rewrite the snippet so that it doesn't require to use a bunch of variables, which is a bad idea anyways

score 2 · Answer 2 · answered Oct 29 '09 at 01:41

2

This also works, but will likely make anyone you might happen to work with frustrated as its very unintuitive:

// pass variables by string name
extract(array_map('trim', compact('a', 'b', 'c')));

answered Oct 29 '09 at 01:41

ken

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score 1 · Answer 3 · answered Oct 29 '09 at 01:12

1

I don't think you can pass a variable-length list of args by reference.

You could pass in an array of references.

function trimAll($array) {
    foreach($array as $k => $v) {
        if(isset($array[$k]) && is_string($array[$k]))
            $array[$k]=&trim($array[$k]);
    }
}

... and suitably modify your call to create an array of references.

$a='  aaa ';
$b='  bbb ';
$c='  ccc ';
trimAll(array(&$a,&$b,&$c));

answered Oct 29 '09 at 01:12

martin clayton

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If you add prefix the argument $array in the function trimAll with &, will that then enable you to use it normally – alex Oct 29 '09 at 01:15
It won't, it complains Fatal error: Only variables can be passed by reference on line 15 http://codepad.org/FItzMVjn – bobo Oct 29 '09 at 01:22
1

maybe I should adopt approach's stereofrog – bobo Oct 29 '09 at 01:33
@ken: this is not call-time pass by reference. putting references in an array is specifically allowed – newacct Oct 07 '11 at 09:16

score 1 · Answer 4 · edited Oct 07 '11 at 17:59

1

I'm not convinced that this is possible using func_get_args, though a comment on it's PHP manual page suggests one possible alternative solution: http://uk3.php.net/manual/en/function.func-get-args.php#90095

However user187291's workaround looks far simpler.

edited Oct 07 '11 at 17:59

Shog9

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answered Oct 29 '09 at 01:25

PeterJCLaw

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yes, but it requires call-time pass-by-reference, which is deprecated – newacct Oct 07 '11 at 09:21

score 0 · Answer 5 · answered Oct 29 '09 at 00:55

0

Have you tried passing in the variables by Reference.

trimAll(&$a,&$b,&$c)

answered Oct 29 '09 at 00:55

Toby Allen

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Call-time pass-by-reference (what you suggest) is deprecated in PHP 5. – ken Oct 29 '09 at 01:27
I know its depreciated, but is it not also about the only way he can do what he wants unless there is a way of declaring that all (unknown number of) properties of a method will be passed byref – Toby Allen Oct 29 '09 at 17:20
This will still not work with the `trimAll` function as it is written (have you tried it?) – newacct Oct 07 '11 at 09:24

score 0 · Answer 6 · answered Oct 07 '11 at 09:28

This works, but uses call-time pass-by-reference, which is deprecated in PHP 5.3:

function trimAll() {
    $backtrace = debug_backtrace();
    foreach($backtrace[0]['args'] as &$arg)
        if(isset($arg) && is_string($arg))
            $arg=trim($arg);
}
$a='  aaa ';
$b='  bbb ';
$c='  ccc ';
trimAll(&$a,&$b,&$c);
echo "$a\n";
echo "$b\n";
echo "$c\n";

Trimming string variables by reference in PHP5

6 Answers6