#variable scope

Part 1 of our homage to John Wick 2, in theaters February 10.

Director Jon Chang

Talent Nea Dune

Shutter Stephen Ciuccoli and Jon Chang

Scope of variables when function in function in JavaScript

Variables in a function only affect the function in JavaScript, can’t be used in another functions, as we know. But how about the scope of variable when function in function. (more…)

The issue

Recently, a colleague posted the following code snippet and asked for an explanation of why it didn't work as they expected:

x = [] for i in range(1, 10): x.append(lambda z: "%d" % i + z) print(list(x[i]("banzaii") for i in range(0, 9)))

which results in:

['9banzaii', '9banzaii', '9banzaii', '9banzaii', '9banzaii', '9banzaii', '9banzaii', '9banzaii', '9banzaii']

instead of the expected:

['1banzaii', '2banzaii', '3banzaii', '4banzaii', '5banzaii', '6banzaii', '7banzaii', '8banzaii', '9banzaii']

The explanation

The key is that the variable i is outside of the scope of the created lambda. Only the variable z is in the lambda scope. Essentially it would be like writing:

x=[] def fn(z): return "%d" % i + z for i in range(1, 10): x.append(fn)

which makes it a lot more obvious what is going on. The global variable i is used each time the function is actually called--which is after the for loop is executed and i is now 9.

The fix

The fix is simple: make sure that i is in the scope of the function that is appended to array x.

Solution 1: using a closure

def makefunc(i): def inner(z): return "%d" % i + z return inner x=[] for i in range(1, 10): x.append(makefunc(i))

In this example, a closure around the function inner passes i into the returned function's scope.

Solution 2: using lambda to create an anonymous closure

x=[] for i in range(1, 10): x.append((lambda n: lambda z: "%d" % n + z)(i))

Here, we are appending the result of calling a lambda function with the argument i which returns a lambda that uses that value. This is exactly equivalent to the previous example, but for people who like one-liners. :)

Solution 3: using a lambda with a default argument

x=[] for i in range(1, 10): x.append(lambda z, n=i: "%d" % n + z)