What will be the output of the following code?
def outer_function(x):
def inner_function(y):
nonlocal x
x += y
return x
return inner_function
add_5 = outer_function(5)
result = add_5(3)
print(result)
a)
5
b)
3
c)
8
d)
13
Correct answer is option 'C'. Can you explain this answer?

Question

What will be the output of the following code?
def outer_function(x):
def inner_function(y):
nonlocal x
x += y
return x
return inner_function

add_5 = outer_function(5)
result = add_5(3)

print(result)

a)
5
b)
3
c)
8
d)
13

Correct answer is option 'C'. Can you explain this answer?

Answer

The outer_function defines a local variable x and an inner function inner_function that modifies x by adding y to it. The outer_function returns the inner_function. When outer_function(5) is called, it returns the inner_function, which is assigned to add_5. When add_5(3) is called, it adds 3 to the current value of x (which is 5) and returns 8.

Answer

The output of the code will be 8.

Explanation:
- The code defines a function called `outer_function` that takes a parameter `x`.
- Inside `outer_function`, there is another function called `inner_function` defined. This inner function takes a parameter `y`.
- The `nonlocal` keyword is used to indicate that the `x` variable inside `inner_function` is the same as the `x` variable in the outer function scope.
- Inside `inner_function`, the value of `x` is updated to the value of `y`.
- The `inner_function` is then returned from `outer_function`.
- The line `add_5 = outer_function(5)` calls `outer_function` with an argument of 5 and assigns the returned `inner_function` to the variable `add_5`.
- Finally, the line `result = add_5(3)` calls `add_5` (which is the `inner_function`) with an argument of 3 and assigns the returned value to the variable `result`.
- Since the value of `x` in `inner_function` is updated to 3, and it is returned as the result, the output will be 8.

In summary, the code defines a closure where the inner function has access to the variable in the outer function and updates its value. The returned inner function is then called with an argument, resulting in the updated value being returned as the output.

Similar EmSAT Achieve Doubts

What will be the output of the following code?def outer():x = 5 def inner(): nonlocal x x = 10 inner() print(x)outer()

What will be the output of the following code?def outer_function():x = 10 def inner_function(): nonlocal x x += 5 print(x) inner_function() print(x)outer_function()

What will be the output of the following code?#include <iostream>int myFunction(int x) { return x * 2;}int myFunction(int x, int y) { return x + y;}int main() { int result1 = myFunction(2); int result2 = myFunction(2, 3); std::cout << result1 << " " << result2 << std::endl; return 0;}

What will be the output of the following code?#include <iostream>int myFunction(int x) { return x * x;}int myFunction(int& x) { return ++x;}int main() { int value = 5; int result = myFunction(valu e); std::cout << result << std::endl; return 0;}

What will be the output of the following code?#include <iostream>int myFunction(int x) { if (x <= 1) return x; return myFunction(x - 1) + myFunction(x - 2);}int main() { int result = myFunction(5); std::cout << result << std::endl; return 0;}

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