What does this mean: $foo = new $foo();?

Question

I've never seen something like this before.

$dbTable = new $dbTable();

We are storing an Object Instance inside $dbTable ?

Are we transforming a string into an object ?

Here's the context:

protected $_dbTable;

    public function setDbTable($dbTable)
    {
        if (is_string($dbTable)) {
            $dbTable = new $dbTable();
        }
        if (!$dbTable instanceof Zend_Db_Table_Abstract) {
            throw new Exception('Invalid table data gateway provided');
        }
        $this->_dbTable = $dbTable;
        return $this;
    }

From php manual here: http://www.php.net/manual/en/language.oop5.basic.php

We can read:

If a string containing the name of a class is used with new, a new instance of that class will be created. If the class is in a namespace, its fully qualified name must be used when doing this.

But this seems to be a concatenation operation between a string and those "things": () - without using a dot. So I'm still not sure about what's going on here.

Answer 1

No, the line:

if (is_string($dbTable)) {

Means that a new $dbTable will only be instantiated if the input is a string. So what is going on here is that $dbTable contains the name of a class that is created when that code executes:

$dbTable = new $dbTable();

And then later on the code checks to make sure an object of the proper type (Zend_Db_Table_Abstract) is created. As Stefan points out, an instance of the class can be passed directly, in which case the code you mentioned is not even executed.