Karl Kofnarson wrote: > > Karl, > > > > Usually when using this idiom, fun_basket would return a tuple of all of the > > defined functions, rather than one vs. the other. So in place of: > >> if f == 1: > >> return f1 > >> if f == 2: > >> return f2 > > Just do > >> return f1, f2 > > (For that matter, the argument f is no longer needed either.) > > > > Then your caller will get 2 functions, who share a common var. You don't > > call fun_basket any more, you've already created your two "closures". Call > > fun_basket using something like: > > > > z1,z2 = fun_basket(None) > > > > And then call z1() and z2() at your leisure - they should have the desired > > behavior. > > > > -- Paul > > Thanks a lot Paul and for the other answers. The things are now > clear to me. In fact, in the Lisp example that I mentioned, you > get a list (or let it be association list) of the internal > functions. Then you can call them separately and they work as > you expect but it's due to the fact only that you got them created > at the same time.
I played around a bit. The following is a 'borg' version in that there is only one counter shared between all calls of the outer function: >>> def fun_borg_var(initial_val=0): ... def borg_var_inc(x=1): ... fun_borg_var._n += x ... return fun_borg_var._n ... def borg_var_dec(x=1): ... fun_borg_var._n -= x ... return fun_borg_var._n ... try: ... fun_borg_var._n = fun_borg_var._n ... except: ... fun_borg_var._n = initial_val ... return (borg_var_inc, borg_var_dec) ... >>> up1, dn1 = fun_borg_var() # get an inc/decrementer >>> up1(0) 0 >>> up1() 1 >>> up1() 2 >>> dn1() 1 >>> dn1() 0 >>> dn1() -1 >>> up2, dn2 = fun_borg_var() # get another inc/decrementer >>> up2(0) # looks like the same _n -1 >>> up2(3) 2 >>> up1(3) 5 >>> - Paddy. -- http://mail.python.org/mailman/listinfo/python-list
