PCEP Computer Programming and Python Fundamentals Practice Question
Consider code:
def outer():
x = 1
def inner():
nonlocal x x = 2 inner()
print(x)
outer() What is printed?
⚠ Common exam trap
The PCEP exam often tests the distinction between `nonlocal` and `global`, and the trap here is that candidates mistakenly think `nonlocal` is unnecessary or causes an error, or they assume the inner assignment creates a separate local variable that does not affect the outer scope.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
2
The `nonlocal` declaration inside `inner()` binds the variable `x` to the `x` defined in the enclosing `outer()` function. When `inner()` assigns `x = 2`, it modifies that outer `x`, so after `inner()` returns, `print(x)` in `outer()` outputs 2.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
2
Why this is correct
The `nonlocal x` declaration binds `inner`'s assignment to the enclosing `outer` function's variable, not a new local one. So `x = 2` mutates the same cell that `print(x)` reads after `inner()` returns, satisfying the stem's requirement that the enclosing scope's value be updated. Output is 2.
- ✗
1
Why it's wrong here
nonlocal x makes inner's assignment mutate outer's variable, so 2 is printed, not 1. Printing 1 would require inner to omit nonlocal and create its own local x, leaving outer's x untouched at its initial value.
- ✗
Error
Why it's wrong here
The nonlocal declaration rebinds outer's x to 2, so no error occurs; the call prints 2. Error would arise only from referencing an unbound name, such as declaring global x without an existing module-level binding, or assigning before nonlocal.
- ✗
None
Why it's wrong here
The function prints 2: nonlocal rebinds the enclosing x, and outer() returns None only as its own value, which is never printed. None would appear solely if the code printed the call's result, as in print(outer()).
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