PHP 5.3 pass function by reference
<p>Is it possible to pass functions by reference?</p>

<p>Something like this:</p>

<pre><code>function call($func){
    $func();
}

function test(){
    echo "hello world!";
}

call(test);
</code></pre>

<p>I know that you could do <code>'test'</code>, but I don't really want that, as I need to pass the function by reference.</p>

<p>Is the only way to do so via anonymous functions?</p>

<p>Clarification: If you recall from C++, you could pass a function via pointers:</p>

<pre><code>void call(void (*func)(void)){
    func();
}
</code></pre>

<p>Or in Python:</p>

<pre><code>def call(func):
    func()
</code></pre>

<p>That's what i'm trying to accomplish. </p>
