Functions within functions, closures, and variable scopes in Python
Why Stack Overflow recommends it
Stack Overflow users link it when answering questions about nested functions, nonlocal, and why an inner function sees an outer variable, calling it the most complete explanation of Python variable scope.
Recommended by 4 people across 5 different months, 2022 to 2024.
What it is
mCoding's James Murphy walks through Python's scoping rules for nested functions in seven escalating 'levels', from returning a global to conditional locals, nonlocal, cells and closures, lambdas and comprehensions. Each level uses short code examples to show how Python decides at compile time where a name is looked up.
What people said
Strong rec quamrana · 2023-11 · Stack Overflow · on “Why is the nonlocal assignment of this variable not considered a previous binding?”
This video [this video] by mcoding [LINK] is the ultimate in explaining all the variations of the scope of variables around nested functions.
Recommends quamrana · 2022-11 · Stack Overflow · on “Python variable scoping in nested function calls”
Yes, I think that in this case you will need `nonlocal res` just inside `def s(root):`. This extensive video [this video] by mCoding explains the scoping rules in python.
Recommends matszwecja · 2022-12 · Stack Overflow · on “How does closures see context variables into the stack?”
There is no universal mechanism. I suggest watching recent mCoding video [this video] on that very topic in Python. In short, at compile time, Python decides in what scope the variable will be looked up. In our case, since `x` is not assigned to in `b`, we check an outer function - it is there, so Python "knows" to look up value of `x` in `a` whenever we need it in `b`. Then, when we actually reach call to `b` Python looks it up in `a` scope.
Recommends zaki98 · 2023-01 · Stack Overflow · on “Variable decleration in a 'try except' block of a recursive function”
it won't print t because t is a local variable to each function call and is not recognized in the context of the other calls to d . if you want to use it you either have to make it global or pass it as an argument def d(t=None): try: print(t.n) except: t=c(5) d(t=t) d() More explanation (There is a great video [this video] about this by mCoding i recommend you watch, and the content of the channel is pretty good) Why is it local in d but not b ? What happens is: at compile time python will look at your function and see is t defined anywhere in your function? if yes (the case of d , even if the variable is assigned after it's usage), it will take the value from that scope, if not it will try findi
Recommends Chris · 2024-06 · Stack Overflow · on “Positional arguments in nested functions”
This question relates to this video: [this video] The piece I'm missing is somewhat simpler than the concept the video is covering overall. In the example code below: Why does the inner function (inner()) pick up the value of the argument that is passed in function call of f and g? (i.e. f("y arg"), g("other y arg")) To my untrained eye, the level_five function only has one parameter defined ('n') which is referenced in the variable assignment to 'z' under the level_five function definition. At no point is a parameter called 'y' defined in the outer (level_five) function definition to be passed to inner()....so how does inner() implicitly pick up the passed value as it's y parameter when the oute