no, it has nothing to do with "i" being global. >>> tuple(lambda: i for i in range(10))[0]() 9 >>> tuple(lambda: i for i in range(10))[1]() 9
what i see as a bug is this code not working as expected: >>> def make_foos(names): ... funcs = [] ... for n in names: ... def foo(): ... print "my name is", n ... funcs.append(foo) ... return funcs ... >>> foos = make_foos(["hello", "world", "spam", "bacon"]) >>> foos[0]() my name is bacon >>> foos[1]() my name is bacon >>> foos[2]() my name is bacon >>> i have to create yet another closure, make_foo, so that the name is correctly bound to the object, rather than the frame's slot: >>> def make_foo(name): ... def foo(): ... print "my name is", name ... return foo ... >>> def make_foos(names): ... return [make_foo(n) for n in names] ... >>> foos = make_foos(["hello", "world", "spam", "bacon"]) >>> foos[0]() my name is hello >>> foos[1]() my name is world >>> foos[2]() my name is spam -tomer On Jan 24, 2:46 am, "Terry Reedy" <[EMAIL PROTECTED]> wrote: > "gangesmaster" <[EMAIL PROTECTED]> wrote in messagenews:[EMAIL PROTECTED] > | so this is why [lambda: i for i in range(10)] will always return 9. > > No, it returns a list of 10 identical functions which each return the > current (when executed) global (module) variable i. Except for names, > 'lambda:i' abbreviates 'def f(): return i'. > > >>> a=[lambda: i for i in range(10)] > >>> i=42 > >>> for j in range(10): print a[j]()42 > 42 > 42 > 42 > 42 > 42 > 42 > 42 > 42 > 42 > > >>> for i in range(10): print a[i]()0 > 1 > 2 > 3 > 4 > 5 > 6 > 7 > 8 > 9 > > >>> del i > >>> for j in range(10): print a[j]()Traceback (most recent call last): > File "<pyshell#23>", line 1, in -toplevel- > for j in range(10): print a[j]() > File "<pyshell#8>", line 1, in <lambda> > a=[lambda: i for i in range(10)] > NameError: global name 'i' is not defined > > | imho that's a bug, not a feature. > > The developers now think it a mistake to let the list comp variable 'leak' > into the global scope. It leads to the sort of confusion that you > repeated. In Py3, the leak will be plugged, so one will get an exception, > as in the last example, unless i (or whatever) is defined outside the list > comp. > > Terry Jan Reedy -- http://mail.python.org/mailman/listinfo/python-list
