PHP &$string - What does this mean?

The & states that a reference to the variable should be passed into the function rather than a clone of it.

In this situation, if the function changes the value of the parameter, then the value of the variable passed in will also change.

However, you should bear in mind the following for PHP 5:

• Call time reference (as you have shown in your example) is deprecated as of 5.3
• Passing by reference when specified on the function signature is not deprecated, but is no longer necessary for objects as all objects are now passed by reference.

You can find more information here: http://www.php.net/manual/en/language.references.pass.php

And there's a lot of information here: Reference — What does this symbol mean in PHP?

An example of the behaviours of strings:

function changeString( &$sTest1, $sTest2, $sTest3 ) {
    $sTest1 = 'changed';
    $sTest2 = 'changed';
    $sTest3 = 'changed';
}

$sOuterTest1 = 'original';
$sOuterTest2 = 'original';
$sOuterTest3 = 'original';

changeString( $sOuterTest1, $sOuterTest2, &$sOuterTest3 );

echo( "sOuterTest1 is $sOuterTest1\r\n" );
echo( "sOuterTest2 is $sOuterTest2\r\n" );
echo( "sOuterTest3 is $sOuterTest3\r\n" );

Outputs:

C:\test>php test.php

sOuterTest1 is changed
sOuterTest2 is original
sOuterTest3 is changed

You are assigning that array value by reference.

passing argument through reference (&$) and by $ is that when you pass argument through reference you work on original variable, means if you change it inside your function it's going to be changed outside of it as well, if you pass argument as a copy, function creates copy instance of this variable, and work on this copy, so if you change it in the function it won't be changed outside of it

Ref: http://www.php.net/manual/en/language.references.pass.php