Question

def outer():
x = []
def inner(val):
x.append(val)
return x
return inner
f1 = outer()
f2 = outer()
print(f1(10)) \# line P
print(f1(20)) \# line Q
print(f2(30)) \# line R
print(f1(40)) \# line S
Which of the following is/are correct?

• A. Output of line Q is [10, 20]
• B. Output of line S is [10, 20, 40]
• C. Output of line R is [10, 20, 30]
• D. f1 and f2 share the same list x

Hint:

In Python, closures capture variables from their enclosing scope. Each time you call the outer function, you create a new closure with a fresh set of captured variables. This is a powerful feature for creating "stateful" functions or factory functions.

Answer 1

The Correct Option is A, B

Step 1: Understanding the Question:
The question presents a Python code snippet involving nested functions (closures) and asks to predict the output at various lines and describe the relationship between the created function objects.
Step 2: Key Concepts - Python Closures and Scope:
- Closure: When a nested function (like `inner`) refers to a variable from its enclosing scope (the `x` in `outer`), it creates a closure. The `inner` function "remembers" the variable `x` from the `outer` function's scope, even after `outer` has finished executing.
- Separate Scopes: Each call to the `outer()` function creates a *new*, independent scope. This means each call creates its own separate `x` list. Therefore, the list `x` used by `f1` is different from the list `x` used by `f2`.
Step 3: Tracing the Code Execution:
- f1 = outer(): This call creates an instance of the `inner` function. This instance of `inner` has a closure over a list `x = []`. Let's call this `x1`. So, `f1` is a function that appends to `x1`.
- f2 = outer(): This is a *second* call to `outer`. It creates a completely separate instance of `inner`. This new instance has a closure over a *new* list `x = []`. Let's call this `x2`. `f2` is a function that appends to `x2`. `x1` and `x2` are different lists in memory.
- print(f1(10)) \# line P: `f1` is called with 10. It appends 10 to its list `x1`, so `x1` becomes `[10]`. It returns `x1`. Output: `[10]`.
- print(f1(20)) \# line Q: `f1` is called again with 20. It appends 20 to the *same* list `x1`, so `x1` becomes `[10, 20]`. It returns `x1`. Output: `[10, 20]`.
- print(f2(30)) \# line R: `f2` is called with 30. It appends 30 to *its own* list `x2`, so `x2` becomes `[30]`. It returns `x2`. This does not affect `x1`. Output: `[30]`.
- print(f1(40)) \# line S: `f1` is called again with 40. It appends 40 to its list `x1`, so `x1` becomes `[10, 20, 40]`. It returns `x1`. Output: `[10, 20, 40]`.
Step 4: Evaluating the Options:
- (A) Output of line Q is [10, 20]: Our trace shows this is TRUE.
- (B) Output of line S is [10, 20, 40]: Our trace shows this is TRUE. (The provided solution in the PDF incorrectly selects this, it should be (b) Output of line S is [10, 20, 40]. Correcting the option label).
- (C) Output of line R is [10, 20, 30]: Our trace shows the output is `[30]`. So, this is FALSE.
- (D) f1 and f2 share the same list x: As explained, each call to `outer()` creates a new scope and a new list `x`. So, they do not share the list. This is FALSE.
Step 5: Final Answer:
The correct options are (A) and (B). The provided PDF answer key has a small typo, marking the correct description of S's output as (b), which we interpret as (B).