Question

This code will be part of a program that will check if a number is prime or not. I know it’s not particularly elegant, but I want to get it working simply for experience. I think that the function is failing because the logic on the if/elif is wrong, when I run this code, it seems to just go straight to the else clause. Is this a syntax problem, or am I not allowed to do logic checks in if clauses?

list = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

def find_prime(list, n):
    if n in list == False:
        list.append(n)
        print "I'ts in there now."
    elif n in list == True:
        print "It's in there already."
    else:
        print "Error"

find_prime(list, 3)
find_prime(list, 51)

Answer 1

1. list is a bad name for a variable. It masks the built-in list.

2. if n in list == True: doesn’t do what you await: 1 in [0, 1] == True returns False (because, as @Duncan notes, 1 in [0,1] == True is shorthand for 1 in [0,1] and [0,1] == True). Use if n in li: and if n not in li:

3. No reason for an Error, since an element is in the list or it is not in the list. Anything else is programming error.

So your code could look like this:

li = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

def find_prime(li, n):
    if n in li:
        print "It's in there already."
    else:
        li.append(n)
        print "It's in there now."