How do variable functions in PHP work?

Question

In php.net the following is written:

Variable functions won't work with language constructs such as echo, print, unset(), isset(), empty(), include, require and the like. Utilize wrapper functions to make use of any of these constructs as variable functions.

source

What does that mean?
Could anyone give examples because I’ve tried using the variable function in an echo and it worked perfectly:

function city()
{
    return "new york";
}
$var = "city";
echo "city:  "  . $var();

Sharanya Dutta · Accepted Answer · 2014-03-05 00:33:45Z

11

It means you can’t do something like this:

$var = "echo";
$var "Hello World!";

edited Mar 5, 2014 at 0:33

answered Jan 28, 2014 at 15:07

Sharanya Dutta

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2 Comments

qwertynl Over a year ago

+1 Ding ding ding! :-D

CompuChip Over a year ago

The comment about wrapper functions means you /can/ do something like this: function myecho($str) { echo $str; } $var = "myecho"; $var("Hello world!");

oleksii.svarychevskyi · Accepted Answer · 2014-01-28 15:10:06Z

0

You cannot do that with these functionns echo, print, unset(), isset(), empty() because actualy in php they are not functions, they are reserved keywords with functions call like.

answered Jan 28, 2014 at 15:10

oleksii.svarychevskyi

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Comments

nilesh · Accepted Answer · 2014-01-28 15:15:12Z

-1

Correct Way is

function city()
{
    return "new york";
}
$var = "city";
echo "city:  "  . city();

It will Return :

city: new york

answered Jan 28, 2014 at 15:15

nilesh

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1 Comment

phpisuber01 Over a year ago

This answer is not helpful and you've completed negated the purpose in setting $var.

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