Re: Re: Closures and __FUNCTION__

I suppose it's an issue of cloning. Perhaps there's some difference between a cloned closure and a referenced closure? Thanks, Justin Martin Peter Danenberg wrote: >> First, we define $foo and load it with NULL so that it is available for >> referencing. > > It turns out loading $foo is superfluous; I can get away with just: > > $foo = function($foo) use (&$foo) { > $foo(); > } > >> Next, in terms of program logic, we create a closure with a lexical >> ('use') variable of a reference to $foo, which is then assigned to >> $foo. Thus, the reference to $foo in the closure declaration now >> points to the closure itself. > > That much is clear; but why $foo is all of the sudden bound within the > closure when I use by reference (&$foo) as opposed to use by value > ($foo) is mysterious.