Simple function problem, local variable referenced before assignment

Question

I'm very new to coding and am having trouble with functions. What is wrong with this and how can I do this properly?

x = 0
def function():
    x = x + 1
    return(x)
x = 2
x = function()
print(x)

Answer 1

Change the function to this if you want to assign variable located outside the function:

def function():
    global x
    x = x + 1
    return x

Answer 2

If you modify (bind) a variable anywhere in a function, it's considered to be local everywhere in that function unless explicitly declared global or nonlocal^(a). This is covered by the Python execution model:

If a name is bound in a block, it is a local variable of that block, unless ...

In other words, what you have is effectively:

x = 0
def function():
    local_x = local_x + 1 # use of unitialised local_x
    return(local_x)

That means you need something like:

x = 0
def function():
    global x
    x = x + 1 # use of initialised x
    return(x)

---

However, keep in mind that global variables are mostly the wrong way to do things. They're probably okay for small programs but can become a nightmare when you get to a certain complexity level.

One method to do this is to just pass in the variable and pass back the modified value, re-assigning it in the caller:

def function(value):
    return value + 1

def caller():
    non_global_x = function(non_global_x)

Note that non_global_x may still be a global in that case but it can be also any other suitable variable, such as the member of an object.

---

^(a) The difference between these two is that global means global, meaning it will be available everywhere. Non-local variables simply refer to ones somewhere up the hierarchy, but you may find them before you get to global scope (often used in nested functions).