``python def any(): var = 1 print(var + 1, end='') var = 1 any() print(var) `` What is the output of the following snippet?

Question

def any():
 var = 1
 print(var + 1, end='')

var = 1
any()
print(var)

Options

• A22
• B21
• C12
• D11

• A
  3% (1)
• B
  76% (22)
• C
  14% (4)
• D
  7% (2)

Explanation

Option B (21) is correct because Python treats var inside any() as a local variable - it shadows the global var but never modifies it. The function prints var + 1 → 1 + 1 = 2 with end='' (no newline), then print(var) outside the function reads the unchanged global var = 1 and prints 1 with a newline, yielding 21.

• A (22) is wrong because the global var is never reassigned; it remains 1, not 2.
• C (12) is wrong because execution order matters - the function call happens first, printing 2, then the global print adds 1.
• D (11) is wrong because var + 1 evaluates to 2, not 1; addition is performed, not ignored.

Memory tip: Think "local = locked in." A variable assigned inside a function stays inside that function - it can't reach out to change the global, and its changes die when the function returns. If you don't see the global keyword, assume the function's copy is its own.

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