nonlocal keyword in Python 2.x

I'm trying to implement a closure in Python 2.6 and I need to access a nonlocal variable but it seems like this keyword is not available in python 2.x. How should one access nonlocal variables in closures in these versions of python?

Inner functions can read nonlocal variables in 2.x, just not rebind them. This is annoying, but you can work around it. Just create a dictionary, and store your data as elements therein. Inner functions are not prohibited from mutating the objects that nonlocal variables refer to.

To use the example from Wikipedia:

def outer():
    d = {'y' : 0}
    def inner():
        d['y'] += 1
        return d['y']
    return inner

f = outer()
print(f(), f(), f()) #prints 1 2 3

The following solution is inspired by the answer by Elias Zamaria, but contrary to that answer does handle multiple calls of the outer function correctly. The "variable" inner.y is local to the current call of outer. Only it isn't a variable, since that is forbidden, but an object attribute (the object being the function inner itself). This is very ugly (note that the attribute can only be created after the inner function is defined) but seems effective.

def outer():
    def inner():
        inner.y += 1
        return inner.y
    inner.y = 0
    return inner

f = outer()
g = outer()
print(f(), f(), g(), f(), g()) #prints (1, 2, 1, 3, 2)

Rather than a dictionary, there's less clutter to a nonlocal class. Modifying @ChrisB's example:

def outer():
    class context:
        y = 0
    def inner():
        context.y += 1
        return context.y
    return inner

Then

f = outer()
assert f() == 1
assert f() == 2
assert f() == 3
assert f() == 4

Each outer() call creates a new and distinct class called context (not merely a new instance). So it avoids @Nathaniel's beware about shared context.

g = outer()
assert g() == 1
assert g() == 2

assert f() == 5

I think the key here is what you mean by "access". There should be no issue with reading a variable outside of the closure scope, e.g.,

x = 3
def outer():
    def inner():
        print x
    inner()
outer()

should work as expected (printing 3). However, overriding the value of x does not work, e.g.,

x = 3
def outer():
    def inner():
        x = 5
    inner()
outer()
print x

will still print 3. From my understanding of PEP-3104 this is what the nonlocal keyword is meant to cover. As mentioned in the PEP, you can use a class to accomplish the same thing (kind of messy):

class Namespace(object): pass
ns = Namespace()
ns.x = 3
def outer():
    def inner():
        ns.x = 5
    inner()
outer()
print ns.x

nonlocal keyword in Python 2.x

Related

Recent Posts