Can someone explain this output between global and local variables?

Question

https://code.sololearn.com/cOW90lbYgkjH

Answer 1

I may be completely wrong about this but just as a guess, I believe it's because you called x as a variable rather than a string in the second foo function.

So because it is being used in foo when you're trying to see if it's a global variable, it is now also a local variable because it's being "called down" in that space and thus being added to the dictionary for the local function.

Because the locals function returns a dictionary with the variable name as the key in string format, and then the value of the same variable as the value of the key.

So essentially the result of locals in your second foo function is:

{ 'x': 1 }

You can test it out by adding a print statement after the first one in the second foo function: print(locals())

Answer 2

Justice you are completely right

Answer 3

Oma Falk Oh, yay! I'm quite terrible with Python, so I had to do a lot of looking up to try to even make sense of it myself. It was a good challenge and thanks a bunch for the confirmation!

Answer 4

Slick The first half of the code popped up as a question in a Challenge. I didn't understand how it was False False.

Justice Your explanation makes perfect sense. Calling x in foo brings it into foo as a local, even though it was already a local in bar, but not in foo up until it is called.

Answer 5

Lookup "namespaces python"

globals() gives you all the names and values of variables in the global namespace.

locals() (When called within a function or method) gives you all the names and values of variables WITHIN the function or method, i.e. their local namespace

What do you have a question about exactly?

Answer 6

Nh