游戏选择while循环中，如何调用对应函数并输出结果的技术求助

原代码存在的问题 #

• 多余的first = input()语句会导致程序额外等待一次输入，打断正常流程
• else语句无对应if判断，属于语法错误
• 退出询问逻辑位置错误，应放在循环内部，每次操作后询问是否退出
• 未实现根据用户选择调用对应游戏函数的核心逻辑

修正后的完整代码 #

import random

def coin_flip():
    gen_num = random.randint(0, 1)
    if gen_num == 0:
        return "Coin flip: Tails"
    else:
        return "Coin flip: Heads"

def roll_d6():
    gen_num = random.randint(1, 6)
    return f"D6 roll: {gen_num}"

def roll_d20():
    gen_num = random.randint(1, 20)
    return f"D20 roll: {gen_num}"

def pick_card():
    suit = random.randint(0, 3)
    if suit == 0: suit = "Spades"
    elif suit == 1: suit = "Hearts"
    elif suit == 2: suit = "Diamonds"
    else: suit = "Clubs"
    value = random.randint(1, 13)
    if value == 1: value = "Ace"
    elif value == 11: value = "Jack"
    elif value == 12: value = "Queen"
    elif value == 13: value = "King"
    return f"Your card: {value} of {suit}"

print('Welcome to the game arena!\nHere are your options:')
stop = False
while not stop:
    print('\t1) Flip a coin\n\t2) Pick a random playing card')
    print('\t3) Roll a 6-sided dice\n\t4) Roll a 20-sided dice')
    choice = input('What would you like to do? ')
    
    # 根据选择调用对应函数并打印结果
    if choice == '1':
        print(coin_flip())
    elif choice == '2':
        print(pick_card())
    elif choice == '3':
        print(roll_d6())
    elif choice == '4':
        print(roll_d20())
    else:
        print('Invalid Operation')
    
    # 询问是否退出
    user_quit = input('Do you want to quit? (y/n): ')
    if user_quit.lower() == 'y':
        stop = True

关键修改说明 #

1. 移除冗余输入：删掉first = input()，避免流程中断
2. 添加选择判断：通过if-elif-else匹配用户输入的选项，直接调用对应函数并打印返回结果
3. 优化条件判断：将pick_card中的独立if改为elif，减少不必要的判断步骤
4. 调整退出逻辑：将退出询问移至循环内部，确保每次操作后都能触发退出判断，正确终止循环
5. 兼容大小写输入：使用lower()方法，支持用户输入大写Y退出程序

内容的提问来源于stack exchange，提问作者jake hanes