How do lexical closures work?

Python is actually behaving as defined. Three separate functions are created, but they each have the closure of the environment they’re defined in – in this case, the global environment (or the outer function’s environment if the loop is placed inside another function). This is exactly the problem, though – in this environment, i is mutated, and the closures all refer to the same i.

Here is the best solution I can come up with – create a function creater and invoke that instead. This will force different environments for each of the functions created, with a different i in each one.

flist = []

for i in xrange(3):
    def funcC(j):
        def func(x): return x * j
        return func
    flist.append(funcC(i))

for f in flist:
    print f(2)

This is what happens when you mix side effects and functional programming.