whats the difference between the following two codes:

Question

this one gives me recursion error
def collatz(n):
if n == 1:
print(1)
if n%2 == 0:
print(n/2)
collatz(n/2)
else:
print(n*3 + 1)
collatz(n*3 + 1)
collatz(7)

this one works properly:
def collatz(n):
if n == 1:
return 1
if n%2 == 0:
print(n/2)
collatz(n/2)
else:
print(n*3 + 1)
collatz(n*3 + 1)
collatz(7)

according to me in the 1st code when we arrive at 1, it should be just printed and code should be stopped...why is that not happening but it happens when we return 1????

Answer 1

print() doesnt stop your code, it just displays what you tell it to.

when you return a value in a function, the function will not go any further.

Answer 2

you can see the benefit of using the return statement here. When you print, it directly outputs, but when you retrn, you can execute the function again bcz, that function (recurssive) becomes the part of thay return statement. So the output is not completed without that.

Answer 3

thank you Slick and M Tamim

Answer 4

The first code sample is an infinite loop, that will be only terminated when the program exeed the recursion level or when the memory consumption is too high.