Python3骰子游戏开发问题：主函数字典值如何传入掷骰子函数

嘿，我一看你的代码就知道问题出在哪了——你调用roll_dice()的时候根本没把玩家名称传进去，函数里也没地方接这个值，那输入提示里自然没法显示玩家名啦！别担心，咱们几步就能搞定：

关键修改点 #

• 给roll_dice函数添加一个参数，用来接收当前玩家的名称
• 在主函数的循环中，调用roll_dice时把对应的玩家名传递过去
• 顺便修复代码里的两个小bug：多余的player +=1和会导致无限递归的return roll_dice

修改后的完整代码 #

from random import randint

def main():
    player_count = int(input('How many players> '))
    player_dict = {}
    # 循环获取玩家名称，改用range(1, player_count+1)刚好覆盖所有玩家数
    for pl in range(1, player_count + 1):
        player_name = input(f'Player {str(pl)} name> ')
        player_dict[pl] = player_name
    # 遍历字典的值，直接拿到每个玩家的名称传递给roll_dice
    for player_name in player_dict.values():
        roll_dice(player_name)

def roll_dice(player_name):
    # 这里用传入的玩家名称格式化提示文本
    start_rolling = input(f'{player_name} roll dice? y/n> ')
    if start_rolling.lower() == 'y':  # 用lower()同时兼容y和Y的输入，更简洁
        roll = randint(1, 6)
        print(f'{player_name} rolled a {roll}!')
    # 去掉原有的return roll_dice，避免无限递归

if __name__ == "__main__":
    main()

额外优化（可选） #

如果你想让玩家可以重复掷骰子直到选择退出，可以给roll_dice函数加个循环：

def roll_dice(player_name):
    while True:
        start_rolling = input(f'{player_name} roll dice? y/n> ')
        if start_rolling.lower() == 'y':
            roll = randint(1, 6)
            print(f'{player_name} rolled a {roll}!')
        elif start_rolling.lower() == 'n':
            print(f'{player_name} is done rolling.')
            break
        else:
            print('Please enter y or n!')

内容的提问来源于stack exchange，提问作者microthingy