If we write x="local’ inside outer() and use print(x) inside inner(), it works. But when we write x= 5 and try x+=10 inside inner(), it shows error, why this happens?
Enclosure functions
hey @nikhil_0406 ,
its because that , if we directly use print command in inner then python will take the previous value of x to be as global .
But when we try to update it , then it doesn’t do so.
it starts to find what is x and finds its value , hence it is not initialized in its scope, it raises error.
I hope this helps.
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but when we tried printing, that time too the string was not in its scope.
hey @nikhil_0406 ,
yes you are right.
But that’s the mystery , when we just directly printed the value , it got that particular value of x , but when we tried to update it , then it raised the error.
Its just because the rules designed in compilation of python and we can’t do anything in it.
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yeah thanks, I understand now.
Good to hear this .
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