Nested Scopes and Lambdas

Question

def funct():
    x = 4
    action = (lambda n: x ** n)
    return action

x = funct()
print(x(2)) # prints 16

... I don't quite understand why 2 is assigned to n automatically?

Answer 1

n is the argument of the anonymous function returned by funct. An exactly equivalent defintion of funct is

def funct():
    x = 4
    def action(n):
        return x ** n
    return action

Does this form make any more sense?

Answer 2

It's not assigned "automatically": it's assigned very explicitly and non-automatically by your passing it as the actual argument corresponding to the n parameter. That complicated way to set x is almost identical (net of x.__name__ and other minor introspective details) to def x(n): return 4**n.