Php: what's the difference between $var and &$var?

The first example creates a copy of the value, whereas the second uses a reference to the original value. So after the first foreach runs, the original array is still untouched. After the second foreach the original array could have been modified since it was handled by reference.

Some native PHP functions already work this way, such as shuffle() which rearranges the contents of your array. You'll notice that this function doesn't return an array, you just call it:

$myArray = array('foo', 'bar', 'fizz', 'buzz');
shuffle( $myArray );
// $myArray is now shuffled

And it works its magic since it works with the array by reference rather than creating a copy of it.

Then there are functions that don't pass anything by reference but rather deal with a copy of the original value, such as ucwords() which returns the new resulting string:

$myString = "hello world";
$myString = ucwords( $myString );
// $myString is now capitalized

See Passing by Reference.

Jonathan's answers describes it very well. Just for completeness, here are your two examples:

1. Just reading values:

   $my_array = range(0,3);
   foreach ($my_array as $my_value) {
       echo $my_value . PHP_EOL;
   }
   

2. Adding some number to each element (thus modifying each value):

   foreach ($my_array as &$my_value) {
       $my_value += 42;
   }
   

   If you don't use &$my_value, then the addition won't have any effect on $my_array. But you could write the same not using references:

   foreach($my_array as $key=>$value) {
       $my_array[$key] = $value + 42;
   }
   

   The difference is that we are accessing/changing the original value directly with $my_array[$key].

DEMO