Python showing local variable referenced before assignment-CodePudding

I've declared 2 variables the same exact way, but when I'm calling them stack is being referenced fine. But python is not able to reference top. What is the issue here?

def isValid(s):
    top=-1
    stack=[-1]*10005

    def push(x):
        print(top)
        top =1
        stack[top]=x
    push()

print(isValid(input()))

CodePudding user response：

The previous answer is mostly correct. Python considers any variable that you directly change the value of to be a local variable. Since you write top = 1, top is directly modified inside push, and is a local variable to push. The variable stack, on the other hand, is not changed.

I consider the fact that Python considers = operators to cause a variable to become local to be a bit of a bug. But it's too ingrained in the language.

And yes, the correct solution is to use nonlocal.

If you removed the top = 1 line, your code wouldn't work, but top would refer to the outer variable.

CodePudding user response：

Because push is indented inside isValid, it can't change the value of top from its "parent" function.

You can use nonlocal:

def isValid(s):
    top=-1
    stack=[-1]*10005

    def push(x):
        nonlocal top
        print(top)
        top =1
        stack[top]=x