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mole7
mole7

Reputation: 53

Writing a Function to check for Divisibility

I am trying to write a function that checks if an int passed into my method in the main file is divisible by 3 and 5.

I am having trouble because I am not sure what to use to check the condition in my method, since the value is passed in via the method call in the main file.

I am also not sure I am using the % operator correctly to check if the value is divisible by 3 and 5. Any guidance on this would be greatly appreciated.

Main:

from divisibleByPackage.isDivisibleBy import *

count_passed = 0
count_failed = 0
if (is_divisible(15) == True):
print("Test #1 passed")
count_passed = count_passed + 1
else:
    print("Test #1 FAILED")
    count_failed = count_failed + 1
if (is_divisible(1) == False):
    print("Test #2 passed")
    count_passed = count_passed + 1
else:
    print("Test #2 FAILED")
    count_failed = count_failed + 1
if (is_divisible(5) == False):
    print("Test #3 passed")
    count_passed = count_passed + 1
else:
    print("Test #3 FAILED")
    count_failed = count_failed + 1
if (is_divisible(0) == True):
    print("Test #4 passed")
    count_passed = count_passed + 1
else:
    print("Test #4 FAILED")
    count_failed = count_failed + 1

    print(str(count_passed) + " tests passed and " + str(count_failed) + " tests failed")

Function imported from PyDev package:

def is_divisible():
    number1 = 3
    number2 = 5

    if (number1 % == 0 && number2 % == 0)
        return True
    else
        return False

I was expecting the method call in my main to return true or false based on the condition, but Eclipse is saying my syntax is incorrect.

Upvotes: 0

Views: 459

Answers (4)

Bastet
Bastet

Reputation: 36

Other answers explain the wrongs. This code will work:

def is_divisible(number1):

    if (number1 % 3 == 0 and number1 % 5 == 0):
        return True
    else:
        return False


count_passed = 0
count_failed = 0
if (is_divisible(15) == True):
    print("Test #1 passed")
    count_passed = count_passed + 1
else:
    print("Test #1 FAILED")
    count_failed = count_failed + 1
if (is_divisible(1) == False):
    print("Test #2 passed")
    count_passed = count_passed + 1
else:
    print("Test #2 FAILED")
    count_failed = count_failed + 1
if (is_divisible(5) == False):
    print("Test #3 passed")
    count_passed = count_passed + 1
else:
    print("Test #3 FAILED")
    count_failed = count_failed + 1
if (is_divisible(0) == True):
    print("Test #4 passed")
    count_passed = count_passed + 1
else:
    print("Test #4 FAILED")
    count_failed = count_failed + 1

print(str(count_passed) + " tests passed and " + str(count_failed) + " tests failed")

Upvotes: 0

Kobby Owen
Kobby Owen

Reputation: 95

There are three errors in the code :

  1. The function is defined not to take arguments, however in the tests, it is called with an argument.
  2. && is not an operator in Python, and is used instead.
  3. The second operand to the % operator is missing. The % is a binary operator, meaning it works with two operands. Example 3%2

The function should be written as


def is_divisible(number):

    if number % 3 == 0 and number % 5 == 0:
        return True
    else:
        return False

Upvotes: 1

John Gordon
John Gordon

Reputation: 33335

Your function needs to accept a parameter, which is the integer to be tested.

Also there's no need to declare 3 and 5 as separate variables. Just use 3 and 5.

Also also, you don't need to explicitly return true/false. You can return the result of the modulus operation, which is itself true/false.

def is_divisible(num):
    return (num % 3 == 0) and (num % 5 == 0)

Upvotes: 0

Joran Beasley
Joran Beasley

Reputation: 113978

number1 % 3 == 0

means a number is divisible by 3

number1 % 5 == 0

means a number is divisible by 5

 number1 % 3 == 0 and number1 % 5 == 0
 # NOTE it is NOT && for 'and' in python

means it is divisible by both 3 AND 5

 number1 % 3 == 0 or number1 % 5 == 0

means it is divisible by either 3 or 5 or maybe both

 number1 % == 0

is just a syntax error (also && is not correct in python)

Upvotes: 0

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