nested scope in python

Question

My question is regarding the rules of enclosing scope. In the snippet below if x=x is not passed to the header of F2 there will be an error

def f1():
 x = 88
 def f2(x=x): # Remember enclosing scope X with defaults
 print(x)
 f2()
f1()

however I dont understand why "x=x" is not needed in the lambda in the below snippet

def func():
 x = 4
 action = (lambda n: x ** n) # x remembered from enclosing def
 return action
x = func()
print(x(2))

• Martin Tournoij: possible duplicate of Short Description of Python Scoping Rules
• abarnert: I don't think that's a dup (although it's certainly related). Neither the question nor the accepted answer addresses the default parameter value hack at all.
• Padraic Cunningham: Are you sure this is the code you meant to post?

Answer 1

In the snippet below if x=x is not passed to the header of F2 there will be an error

No there won't:

>>> def f1():
... x = 88
... def f2():
... print(x)
... f2()
... 
>>> f1()
88

---

You only need the default parameter value hack when you're trying to pass a value that might get changed later. f2 as I've written it captures the variable x from the enclosing scope, so it will print whatever x happens to be at the time. As you've written it, it captures the current value of the variable x, not the variable itself.

For example:

>>> def f1():
... x = 88
... def f2():
... print(x)
... x = 44
... f2()
... 
>>> f1()
44
>>> def f1():
... x = 88
... def f2(x=x):
... print(x)
... x = 44
... f2()
... 
>>> f1()
88

---

For a very common real-life example of where this difference is important, see Why do lambdas defined in a loop with different values all return the same result? in the official FAQ.