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? — newacct
– newacct, Commented Oct 7, 2011 at 9:20

user187291 · Accepted Answer · 2009-10-29 01:15:39Z

4

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

answered Oct 29, 2009 at 1:15

user187291

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Comments

ken · Accepted Answer · 2009-10-29 01:41:19Z

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, 2009 at 1:41

ken

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Comments

martin clayton · Accepted Answer · 2009-10-29 01:12:17Z

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, 2009 at 1:12

martin clayton

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4 Comments

alex Over a year ago

If you add prefix the argument $array in the function trimAll with &, will that then enable you to use it normally

bobo Over a year ago

It won't, it complains Fatal error: Only variables can be passed by reference on line 15 codepad.org/FItzMVjn

bobo Over a year ago

maybe I should adopt approach's stereofrog

newacct Over a year ago

@ken: this is not call-time pass by reference. putting references in an array is specifically allowed

Community · Accepted Answer · 2023-11-17 20:29:45Z

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: https://www.php.net/manual/en/function.func-get-args.php#90095

However user187291's workaround looks far simpler.

edited Nov 17, 2023 at 20:29

CommunityBot

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answered Oct 29, 2009 at 1:25

PeterJCLaw

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1 Comment

newacct Over a year ago

yes, but it requires call-time pass-by-reference, which is deprecated

Toby Allen · Accepted Answer · 2009-10-29 00:55:46Z

0

Have you tried passing in the variables by Reference.

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

answered Oct 29, 2009 at 0:55

Toby Allen

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3 Comments

ken Over a year ago

Call-time pass-by-reference (what you suggest) is deprecated in PHP 5.

Toby Allen Over a year ago

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

newacct Over a year ago

This will still not work with the trimAll function as it is written (have you tried it?)

newacct · Accepted Answer · 2011-10-07 09:28:44Z

0

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";

answered Oct 7, 2011 at 9:28

newacct

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