Python closure function losing outer variable access

Question

I just learned python @ decorator, it\'s cool, but soon I found my modified code coming out weird problems.

def with_wrapper(param1):
    def dummy_wrapper(fn):
         


        

Answer 1

When Python parses a function, it notes whenever it finds a variable used on the left-hand side of an assignment, such as

param1 = 'new'

It assumes that all such variables are local to the function. So when you precede this assignment with

print param1

an error occurs because Python does not have a value for this local variable at the time the print statement is executed.

---

In Python3 you can fix this by declaring that param1 is nonlocal:

def with_wrapper(param1):
    def dummy_wrapper(fn):
        nonlocal param1
        print param1
        param1 = 'new'
        fn(param1)
    return dummy_wrapper

---

In Python2 you have to resort to a trick, such as passing param1 inside a list (or some other mutable object):

def with_wrapper(param1_list):
    def dummy_wrapper(fn):
        print param1_list[0]
        param1_list[0] = 'new'   # mutate the value inside the list
        fn(param1_list[0])
    return dummy_wrapper

def dummy():
    @with_wrapper(['param1'])   # <--- Note we pass a list here
    def implementation(param2):
        print param2