Question

此函数的目标是计算元音，但是在所有情况下都将执行if语句这是代码：

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)

它必须打印1但它要打印4

Answer 1

问题在于，您正在将char与'a'进行比较，然后仅检查字符串值（如果存在），该值是一个值，在这种情况下始终是真实的。

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)

您需要这样做：

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)

或更清洁的选择：

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)

python即使条件为false也会执行if语句

1 个答案: