UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)

Question

When I try this code: ``` a, b, c = (1, 2, 3) def test(): print(a) print(b) print(c) c += 1 test() ``` I get an error from the `print(c)` line that says: ``` UnboundLocalError: local variable 'c' referenced before assignment ``` in newer versions of Python, or ``` UnboundLocalError: 'c' not assigned ``` in some older versions. If I comment out `c += 1`, both `print`s are successful. I don't understand: why does printing `a` and `b` work, if `c` does not? How did `c += 1` cause `print(c)` to fail, even when it comes later in the code? It seems like the assignment `c += 1` creates a local variable `c`, which takes precedence over the global `c`. But how can a variable "steal" scope before it exists? Why is `c` apparently local here? See also Using global variables in a function for questions that are simply about how to reassign a global variable from within a function, and Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope? for reassign…

Accepted Answer

Python treats variables in functions differently depending on whether you assign values to them from inside or outside the function. If a variable is assigned within a function, it is treated by default as a local variable. Therefore, when you uncomment the line, you are trying to reference the local variable `c` before any value has been assigned to it.

If you want the variable `c` to refer to the global `c = 3` assigned before the function, put

```
global c
```

as the first line of the function.

As for python 3, there is now

```
nonlocal c
```

that you can use to refer to the nearest enclosing function scope that has a `c` variable.

UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)

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