What will be the output of the following code?def outer():x = 5 def in...
Explanation:
The code provided defines a function named "outer" which contains another function named "inner". Inside the "inner" function, the keyword "nonlocal" is used to indicate that the variable "x" is not local to the "inner" function, but is a variable defined in the nearest enclosing scope, which is the "outer" function.
Let's break down the code and the execution flow to understand the output:
1. The "outer" function is called, and it defines a variable "x" with a value of 5.
2. The "inner" function is defined inside the "outer" function.
3. Inside the "inner" function, the keyword "nonlocal" is used to indicate that the variable "x" is not local to the "inner" function but is a variable in the enclosing scope, which is the "outer" function.
4. The value of "x" is changed to 10 inside the "inner" function.
5. The "inner" function is called.
6. After the execution of the "inner" function, the value of "x" is printed.
7. The output of the code is 10.
The reason the output is 10 is that the "nonlocal" keyword is used to indicate that the variable "x" should refer to the same variable in the enclosing scope, rather than creating a new local variable. So when the value of "x" is changed inside the "inner" function, it affects the variable "x" in the "outer" function as well.
If the "nonlocal" keyword was not used, the code would create a new local variable "x" inside the "inner" function, and the output would be 5, as the change inside the "inner" function would not affect the variable in the "outer" function.
Therefore, the correct answer is option B) 10.
What will be the output of the following code?def outer():x = 5 def in...
The outer function defines a variable x and calls the inner function, which uses the nonlocal keyword to modify the outer variable x. After calling inner, the value of x is changed to 10, which is printed as the output.
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