PHP变量函数和语言构造

Question

I'm trying to understand why PHP documentation for variable functions states that:

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.

I tried some of these and they work just fine:

function animal() {

    return 'Monkey';

}

$animal = 'animal';

echo $animal();

returns Monkey - just as one would expect.

Same result with print construct - then I tried unset() and it also works absolutely fine:

function getIndex() {

    return 0;

}

$index = 'getIndex';

$array = array(

    'Monkey',
    'Gorilla'

);

unset($array[$index()]);

print_r($array);

this returns Array ( [1] => Gorilla ) .

Is there something I'm missing here? Just to add - I'm using PHP 5.5.14.

Answer 1

They mean another usage:

<?php
$var = "some variable";
$a = "unset"; //also print, isset, echo, include

// you cannot do this:
$a($var);

Of course you can unset or print variable with string containing function name...

(Or am I missing something? :) )

Answer 2

You're not actually using any language constructs as a variable function:

But try

$function = 'echo'; 
$function('Hello World');

and it won't work, exactly as described in the docs

Use a wrapper function around echo as described in the manual, and then you can use that function as a variable function

function myecho($value) {
    echo $value;
}

$function = 'myecho'; 
$function('Hello World');