On 15/12/2013 16:54, Rafael Knuth wrote:
Hej,
I stumbled upon this program here (Python 3.3.0) and I don't quite
understand how the for loop plays with the return True statement:
def is_prime(number):
for element in range(2, number):
if number % element == 0:
return False
return True
Now, I would expect the following in case I call the function with the
variable 3:
number = 3
for element in range(2, 3):
3 % 2 != 0:
Loop ends and program returns True.
Let's do the same with the variable 9:
number = 9
for element in range(2,9):
3 % 2 != 0:
My assumption is that the program should end the loop after the first
iteration again and it then should return True.
But instead, the program returns False (correctly for obvious reasons
because 9 is not a prime number). Can anyone help me understand what
error in reasoning I am making here?
Thanks!
All the best,
Raf
If you pass 3 into the function the code will only loop for the value 2,
that's simply how range works. Your friend here is the print function
(statement in Python 2, unless you use IIRC from __future__ import
print_function). So try something like this and watch what happens.
def is_prime(number):
print('is_prime number = ', number)
for element in range(2, number):
print('is_prime element = ', element)
if number % element == 0:
print('is_prime return False')
return False
print('is_prime return True')
return True
for i in range(2, 10):
is_prime(i)
print()
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
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