Question

I have a following simple code:

def get():
    return [lambda: i for i in [1, 2, 3]]

for f in get():
    print(f())

As expected from my python knowledge, output is 3 – entire list will contain last value of i. But how this works internally?

AFAIK, python variables are simply reference to objects, so first closure must enclose object first i reference – and this object is definitely 1, not 3 O_O. How it happens that python closure encloses variable itself instead of object this variable reference? Does it save variable name as plain text, some “reference to variable” or what?

Answer 1

Closures don’t refer to variables but rather to scopes. Since the last value of i in its scope is ‘3’, all three closures return the same. To “lock” the current value of a variable, create a new scope just for it:

def get() : return [ (lambda x: lambda: x)(i) for i in [ 1, 2, 3 ] ]
for f in get() : print( f() )