Question

Question:

Define a function isVowel(char) that returns True if char is a vowel (‘a’, ‘e’, ‘i’, ‘o’, or ‘u’), and False otherwise. You can assume that char is a single letter of any case (ie, ‘A’ and ‘a’ are both valid).

Do not use the keyword in. Your function should take in a single string and return a boolean.

Code Given:

def isVowel(char):
    '''
    char: a single letter of any case

    returns: True if char is a vowel and False otherwise.
    '''

My Code:

def isVowel(char):
    '''
    char: a single letter of any case

    returns: True if char is a vowel and False otherwise.
    '''
    if char == 'a' or 'e' or 'i' or 'o' or 'u' or 'A' or 'E' or 'I' or 'O' or 'U':
        return True
    else:
        return False

My Problem:
My output is always True. What am I doing wrong?

Answer 1

Your if statement:

if char == 'a' or 'e' or 'i' or 'o' or 'u' or 'A' or 'E' or 'I' or 'O' or 'U':
        return True

is equivalent to:

if (char == 'a') or 'e' or 'i' or 'o' or 'u' or 'A' or 'E' or 'I' or 'O' or 'U':

which will always be evaluated to either True, or e which is also True, and hence your function always returns True.

Change your if-statement to:

if char == 'a' or char == 'e' or char == 'i' so on...:
        return True

---

But, this problem is really simple if you can use in operator. This goes like this:

def isVowel(char):
    return char.lower() in 'aeiou'