Logic in thos code?

Question

def func(a,b):
s= 0
for i in range(a):
for j in range(b):
s+=(a+b)%3
return s
print(func(3,4))

Answer 1

Loop continues 12 times
(a + b) % 3 = 7%3 = 1
so s = 1*12 = 12

Answer 2

func(3,4)

So, the innermost statement, s+=(3+4)%3 runs for 12 (3*4) times.

(3+4)%3
= 7%3
= 1

s
= 1+1+1+1+1+1+1+1+1+1+1+1
= 12

s is returned.
12 is returned.

print(func(3,4))
print(12)

12

Answer 3

Your print(func(3,4)) command calls the func function with a = 3 and b = 4.

So a+b = 7, and so (a+b)%3 = 1. Your two ‘for’ loops mean that the s+=(a+b)%3 line is executed 3*4 (=12) times, each time incrementing s by 1.

All this means your function will return 12.

Answer 4

AMAN TOMAR Please, as jay told you here https://www.sololearn.com/Discuss/1557917/?ref=app try to not "call" other users in Q&A section... Here there are many users that help then try to wait instead to call determinate users after 5 minutes that you posted the question... If everyone would do it, there would no peace for some users

Answer 5

noted KrOW

Answer 6

AMAN TOMAR 👍👍👍