Answer-In Python, what will be the output of the following code snippet, considering scope rules? x = 5&nb...

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Answer-In Python, what will be the output of the following code snippet, considering scope rules?  x = 5&nb...

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The code demonstrates the use of the nonlocal keyword in Python, which allows a variable defined in an enclosing but not global scope to be modified. Heres a breakdown   The variable x  5 is declared globally.  Inside the func function, a local x  10 is declared.  Within the nested inner function, the nonlocal x statement is used. This tells Python that x refers to the x variable in the enclosing func scope not the global scope.  The x is incremented by 1 x  1, updating it to 11.  The inner function returns x, and func also returns the result from inner. Hence, the final output is 11.   The nonlocal keyword ensures that changes to x in the inner function persist in the func scope. Without it, Python would treat x in inner as a new local variable, causing an UnboundLocalError.   Why Other Options Are Incorrect    Option A 6 This assumes that nonlocal affects the global x. However, nonlocal only modifies variables in the closest enclosing non-global scope. Thus, the global x  5 remains unchanged.   Option C 10 This would be true if no modification occurred to the local x inside the inner function. However, x is incremented by 1 due to nonlocal.   Option D UnboundLocalError This error would occur if nonlocal was omitted and x was incremented without declaration, as Python would not recognize it as being defined in the current scope.   Option E None This would result if func or inner explicitly returned None. Since the function returns x, this is incorrect.

In Python, what will be the output of the following code snippet, considering scope rules? x = 5 def func(): x = 10 def inner(): nonlocal x x += 1 return x return inner() print(func())

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