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Modifying a variable before it's assigned locally

🐛 Fix the bug Topic: functions Difficulty: medium Free

What this problem practises

Function bugs are mostly about names and values: positional argument order, default parameters evaluated once at definition time, local versus global scope, and the implicit None returned whenever a function has no return.

This is a fix-the-bug problem: a short program is given in its broken state, and the task is to repair the code so it produces the required output. The interactive editor runs your fixed version in the browser and checks it for you — no setup, no local interpreter.

Difficulty: medium — expect to hold two interacting rules in your head at once, such as a boundary combined with a mutation.

The challenge

This should increase a counter and print it, but crashes. Fix it.

The code runs in PyDebug’s in-browser compiler. Open the problem, run the snippet, watch the exact error or output, and keep iterating until it is right. Signed-in solves count toward XP and your daily streak.

The error it raises

Run the broken snippet and Python stops with this exact error: UnboundLocalError: local variable 'count' referenced before assignment. Match it to the Python errors guide, then repair the line that produced it — knowing the message is half the fix.

The snippet

The interactive editor pre-fills exactly this code, broken and ready to repair:

count = 0

def increment():
    count += 1
    print(count)

increment()

Understand the rule

If a function assigns to a name anywhere in its body, Python treats that name as local throughout the function. Reading it before the local assignment therefore raises UnboundLocalError, even when a global variable with the same name exists.

How to trace it

  1. Scan the whole function for assignments to the name, not just the line that fails.
  2. Separate the local variable from any global or outer-scope variable with the same spelling.
  3. Pass the starting value as an argument when the function should transform caller data.
  4. Use nonlocal or global only when shared state is truly intentional.

Common wrong fixes

Rule to remember

A local assignment changes the scope decision for the entire function. Explicit parameters are usually clearer than reaching into outer state.

Solve it in the interactive editor →

🧩 Learning path: Functions

Problem 11 of 12 in this path. Work them in order — each one builds on the rule before it.

← Tuple unpacking with functionsNext in path: Mutable default argument trap →

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