Arbitrary scope nesting in Python

Question

The following python example is from here:

def f1():
 x = 99
 def f2():
 def f3():
 print(x)
 print('f3')
 f3()
 f2()
f1()

and this is the output:

99
f3

Why isn't the output the following?

99
f3 # from f1()
99
f3 # from f3()
99
f3 # from f2()

• juanpa.arrivillaga: @lmiguelvargasf I believe the question is really about the call stack, not scope.

Answer 1

It is because as the code is unwrapped it simplifies to:

x = 99
print(x)
print('f3')

This is the sequence of unwrapping:

def f1():
 x = 99
 def f2():
 def f3():
 print(x)
 print('f3')
 f3()
 f2()
f1()

becomes

 x = 99
 def f2():
 def f3():
 print(x)
 print('f3')
 f3()
 f2()

becomes

 x = 99
 def f3():
 print(x)
 print('f3')
 f3()

becomes

 x = 99
 print(x)
 print('f3')

which is then what is executed.

Answer 2

Consider the code:

def f():
 print(123)

which when executed won't print anything. def is only defining the function - it's not executed until we do f().

Similarly your example only defines the inner functions, but only when f3 is called do the print statements execute.

Answer 3

You should consider the Python scope rules (LEGB). PYthon will look for:

L: local variables

E: enclosing variables (variables defined in enclosing functions)

G: global variables (variables at the zero-indent level)

B: built-in variables

How your program is executed:

1. f1 is defined and then f1 is called
2. When f1 is called, x takes the value of 99, then f2 is defined, and then f2 is called.
3. When f2is called, f3 is defined, and then f3 is called.
4. When f3 is called, x is printed, and since there is no local x in f3, it uses the enclosing x that is local to f1. Then, you are printing exactly the string "f3" with the print function.
5. Finally, f3 returns, f2 returns, f1 returns, and your program finishes.