python the if statement is executed even if the condition is false

Question

this function goal is to count the vowels but the if statement is being executed in all the cases this is the code:

def count_vowels(txt):
  count=0
  txt = txt.lower()
  for char in txt:
    
    if char == "a" or "e" or "i" or "o" or "u":
       count = count+1
    
  print(count)

count_vowels(mark)

it has to print 1 but it's printing 4

its never false `if 'e':` is always a true statement, i think you want `if char in "aeiou":` — Joran Beasley, Jul 24 '20 at 18:55

score 1 · Accepted Answer · answered Jul 24 '20 at 18:56

The issue is that you are comparing char to 'a' and then just checking string values if they exists which is a value and will always be truthy in this case.

def count_vowels(txt):
  count=0
  txt = txt.lower()
  for char in txt:
    
    if char == "a" or "e" or "i" or "o" or "u":
       count = count+1
    
  print(count)

count_vowels(mark)

You would need to do:

def count_vowels(txt):
  count=0
  txt = txt.lower()
  for char in txt:
    
    if char == "a" or char == "e" or char == "i" or char == "o" or char == "u":
       count = count+1
    
  print(count)

count_vowels(mark)

Or a cleaner alternative:

def count_vowels(txt):
  count=0
  txt = txt.lower()
  for char in txt:
    
    if char in ['a', 'e', 'i', 'o', 'u']:
       count = count+1
    
  print(count)

count_vowels(mark)

or just `return len([c for c in txt.lower() if c in "aeiou"])` — Samwise, Jul 24 '20 at 19:01

python the if statement is executed even if the condition is false

1 Answers1