python - Why doesn't this closure modify the variable in the enclosing scope?

Question

This bit of Python does not work:

def make_incrementer(start):
    def closure():
        # I know I could write 'x = start' and use x - that's not my point though (:
        while True:
            yield start
            start += 1
    return closure

x = make_incrementer(100)
iter = x()
print iter.next()    # Exception: UnboundLocalError: local variable 'start' referenced before assignment

I know how to fix that error, but bear with me:

This code works fine:

def test(start):
    def closure():
        return start
    return closure

x = test(999)
print x()    # prints 999

Why can I read the start variable inside a closure but not write to it? What language rule is causing this handling of the start variable?

Update: I found this SO post relevant (the answer more than the question): Read/Write Python Closures

Answer 1

Whenever you assign a variable inside of a function it will be a local variable for that function. The line start += 1 is assigning a new value to start, so start is a local variable. Since a local variable start exists the function will not attempt to look in the global scope for start when you first try to access it, hence the error you are seeing.

In 3.x your code example will work if you use the nonlocal keyword:

def make_incrementer(start):
    def closure():
        nonlocal start
        while True:
            yield start
            start += 1
    return closure

On 2.x you can often get around similar issues by using the global keyword, but that does not work here because start is not a global variable.

In this scenario you can either do something like what you suggested (x = start), or use a mutable variable where you modify and yield an internal value.

def make_incrementer(start):
    start = [start]
    def closure():
        while True:
            yield start[0]
            start[0] += 1
    return closure