A.) def outer():
x=“local”
def inner():
print(x)
inner()
print(x)
B.)def outer():
x=10
def inner():
nonlocal x
x+=5
print(x)
inner()
print(x)
In case A we don’t require " nonlocal " keyword in it but in case B we require “nonlocal” keyword . they both are placed at same place in the code . is there any difference for using "nonlocal " keyword for string and inegers .
my question Is why in case A "nonlocal " keyword not used ?
Hey @Coder_sash2123, this is because of the fact that when you use the assignment operator (=) inside any function in python , so the variable associated becomes local to the function. So here in the second case if you don’t specify x as non local, it would give you an error in case of x+=5 , but it would not give the error in case you just print the value of x.
I hope it is clear. If it is not you can ask me again.
Happy coding 
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