Scope of lambda functions and their parameters?

Question

I need a callback function that is almost exactly the same for a series of gui events. The function will behave slightly differently depending on which event has called it.

Answer 1

Python does uses references of course, but it does not matter in this context.

When you define a lambda (or a function, since this is the exact same behavior), it does not evaluate the lambda expression before runtime:

# defining that function is perfectly fine
def broken():
    print undefined_var

broken() # but calling it will raise a NameError

Even more surprising than your lambda example:

i = 'bar'
def foo():
    print i

foo() # bar

i = 'banana'

foo() # you would expect 'bar' here? well it prints 'banana'

In short, think dynamic: nothing is evaluated before interpretation, that's why your code uses the latest value of m.

When it looks for m in the lambda execution, m is taken from the topmost scope, which means that, as others pointed out; you can circumvent that problem by adding another scope:

def factory(x):
    return lambda: callback(x)

for m in ('do', 're', 'mi'):
    funcList.append(factory(m))

Here, when the lambda is called, it looks in the lambda' definition scope for a x. This x is a local variable defined in factory's body. Because of this, the value used on lambda execution will be the value that was passed as a parameter during the call to factory. And doremi!

As a note, I could have defined factory as factory(m) [replace x by m], the behavior is the same. I used a different name for clarity :)

You might find that Andrej Bauer got similar lambda problems. What's interesting on that blog is the comments, where you'll learn more about python closure :)