Op 2005-02-19, jfj schreef <[EMAIL PROTECTED]>: > Carl Banks wrote: >> Ted Lilley wrote: >> >> >>>Unfortunately, it doesn't work. It seems the closure keeps track of >>>the variable fed to it dynamically - if the variable changes after >> [...] >>> >>>At least, that's the explanation I'm deducing from this behavior. >> >> >> And that's the correct explanation, chief. > >> >> It is intended that way. As an example of why that is: consider a >> nested function called "printvars()" that you could insert in various >> places within a function to print out the value of some local variable. >> If you did that, you wouldn't want printvars to print the values at >> the time it was bound, would you? > > Allow me to disagree (and start a new "confused with closures" thread:) > > We know that python does not have references to variables. To some > newcomers this may seem annoying but eventually they understand the > pythonic way and they do without them. But in the case of closures > python supports references! > > Nested functions, seem to do two independent things: > 1) reference variables of an outer local scoope > 2) are functions bound dynamically to constants > > These two are independent because in: > ############## > def foo(): > def nested(): > print x > f = nested > x = 'sassad' > f() > x = 'aafdss' > f() > return f > ################## > > once foo() returns there is no way to modify 'x'! > It becomes a kind of constant.
In this particular case yes. But not in general, what about this: >>> def F(): ... l = [] ... def pop(): ... return l.pop() ... def push(e): ... l.append(e) ... return pop, push ... >>> pop, push = F() >>> push(1) >>> pop() 1 >>> push(2) >>> push(3) >>> pop() 3 >>> pop() 2 -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
