Choose the Correct Answer(s)What will be the output after the following statements?def xyz(): a = 56 xyz() print(a)OptionsNameErrora = 56xyz56
Question
Choose the Correct Answer(s)What will be the output after the following statements?def xyz(): a = 56 xyz() print(a)OptionsNameErrora = 56xyz56
Solution
The correct answer is NameError.
Here's why:
In Python, variables that are defined inside a function are local to that function, meaning they cannot be accessed outside of that function. In the given code, variable 'a' is defined inside the function xyz(). When we try to print 'a' outside of the function, Python doesn't recognize 'a' because it's not defined in the global scope. Hence, it will raise a NameError.
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