``python num = 1 def func(): num = 3 print(num, end=' ') func() print(num) `` What is the output of the following snippet?

Question

num = 1

def func():
 num = 3
 print(num, end=' ')

func()

print(num)

Options

• A33
• B11
• CThe code is erroneous.
• D13
• E31

• B
  15% (3)
• C
  5% (1)
• D
  5% (1)
• E
  75% (15)

Explanation

Option E (3 1) is correct because Python's scoping rules treat num = 3 inside func() as a local variable - it shadows the global num without modifying it. So func() prints 3, and the subsequent print(num) sees the original global value, printing 1.

Why the distractors fail:

• A (33): Assumes the assignment inside func() changes the global num to 3 - it doesn't; without global num, it's a new local variable.
• B (11): Assumes the local num = 3 is ignored and the function reads the global 1 instead - Python reads the local assignment first.
• C (erroneous): The code runs cleanly; no UnboundLocalError occurs because num is assigned before it's read inside func().
• D (13): Reverses the call order - func() executes before the outer print(num), so it must be 3 first, then 1.

Memory tip: Think "LEGB" - Python looks up names Local → Enclosing → Global → Built-in. A bare assignment inside a function always creates a Local variable, leaving the Global untouched unless you explicitly declare global num.

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