PHP references question
 

I suspect I'm missing something very obvious here. Can someone explain the behaviour of this code:

which prints out

Inside
test:12345
anothertest:12345
yetanothertest:

Outside
test:12345
anothertest:
yetanothertest:12345

In other words, within the function, the reference assignment seems to have updated $anothertest but not $yetanothertest, yet having left the function, it's the other way round.

With direct assignments ($A = $B) instead of references ($A =& $B), both of the global variables are updated and correctly displayed inside and outside the function.

What values were you expecting?

Assigns the global variable by to reference to the local variable to test; it's not affecting $yetanothertest in the local scope of the function.

and

Is only affecting the reference of the local variable $anothertest...$anothertest does not have scope after the function call, however, if you assigned a value to the variables as you mentioned...it will affect all the variables that reference that data.

I know I'm not explaining it that well, there's a section in the PHP manual dedicated to explaining references over at php.net.

I was expecting $GLOBALS['abc'] to be identical to global $abc inside the function, and it isn't.

As you say, it's explained in the php help manual:

Warning
If you assign a reference to a variable declared global inside a function, the reference will be visible only inside the function. You can avoid this by using the $GLOBALS array. Think about global $var; as a shortcut to $var =& $GLOBALS['var'];. Thus assigning other reference to $var only changes the local variable's reference.

I didn't get that far on that page.