如何用Python的len函数判断单词长度并控制程序启停？

编程作业要求：使用len()函数判断输入变量original_word的字符数量，若该单词长度不等于5（大于或小于5个字符），程序需调用exit()终止运行。现有如下Python代码，请问如何添加该逻辑？

原代码：

original_word: str = input("Enter a 5-character word: ")
character_one: str = input("Enter a single character: ")
print("Searching for " + character_one + " in " + original_word)
if character_one == original_word[0]:
    print(character_one + " found at index 0")
if character_one == original_word[1]:
    print(character_one + " found at index 1")
if character_one == original_word[2]:
    print(character_one + " found at index 2")
if character_one == original_word[3]:
    print(character_one + " found at index 3")
if character_one == original_word[4]:
    print(character_one + " found at index 4")

counter = 0
for c in original_word:
    if c == character_one:
         counter += 1
if counter == 2:
    print(counter, " instances of " + character_one + " found in " + original_word)
if counter == 1:
    print(counter, " instance of " + character_one + " found in " + original_word)
if counter == 0: 
    print("No instances of " + character_one + " found in " + original_word)

只需在获取original_word输入后，立刻添加长度判断逻辑即可，具体修改如下：

修改后的完整代码：

import sys

original_word: str = input("Enter a 5-character word: ")
# 添加长度校验逻辑
if len(original_word) != 5:
    print("输入单词长度不符合要求，程序终止")
    sys.exit()

character_one: str = input("Enter a single character: ")
print("Searching for " + character_one + " in " + original_word)
if character_one == original_word[0]:
    print(character_one + " found at index 0")
if character_one == original_word[1]:
    print(character_one + " found at index 1")
if character_one == original_word[2]:
    print(character_one + " found at index 2")
if character_one == original_word[3]:
    print(character_one + " found at index 3")
if character_one == original_word[4]:
    print(character_one + " found at index 4")

counter = 0
for c in original_word:
    if c == character_one:
         counter += 1
if counter == 2:
    print(counter, " instances of " + character_one + " found in " + original_word)
if counter == 1:
    print(counter, " instance of " + character_one + " found in " + original_word)
if counter == 0: 
    print("No instances of " + character_one + " found in " + original_word)

关键说明 #

• 校验逻辑必须放在获取original_word之后、获取character_one之前，确保不符合条件时直接终止，不执行后续流程
• 若不想导入sys模块，也可直接使用内置的exit()函数，但调用sys.exit()是更规范的写法
• 可自定义提示信息，让用户明确程序终止的原因

内容的提问来源于stack exchange，提问作者kitkat