剪刀石头布代码打印结果不一致答案

【问题标题】：Rock paper scissors code not consistently printing outcome剪刀石头布代码打印结果不一致
【发布时间】：2021-07-26 07:22:03
【问题描述】：

这是我为一个非常基本的石头剪刀布游戏编写的代码。但是，当我运行代码时。它并不总是打印游戏的结果，即。你赢了！，你输或平。我不太确定为什么会这样。救命！！[1]

start_game = input('Type start to start game:').lower()
import random
def play():
    user = input("'rock' , 'paper', 'scissors' ").lower()
    if user == 'quit':
        print("Game ended :(")
        exit()
    computer = random.choice(['rock', 'paper','scissors'])
    print(computer)
    if user == computer:
        return 'tie'
    elif not user == is_win(user, computer):
        return 'You win!'
    else:
        return 'You lost!'

def is_win(player,opponent):
    if (player == 'rock'and opponent == 'scissors') or (player == 'scissors'and opponent == 'paper') or \
            (player=='paper' and opponent == 'rock'):
        return True
    print(play())

while True:
    if start_game != 'start':
        start_game = input('Sorry i dont understand please try again:')
    else:
        print(play())

输出：输入 start 开始游戏：srt 对不起，我不明白，请再试一次：开始 “摇滚”、“纸”、“剪刀”摇滚纸

【问题讨论】：

标签： python

【解决方案1】：

决定获胜者的逻辑有点错误。对象user 只包含用户的选择。函数is_win 应该返回一个布尔值。当用户获胜时，您确实返回True，但当他失败时，您正在调用原始调用函数。

start_game = input('Type start to start game:').lower()
import random
def play():
    user = input("'rock' , 'paper', 'scissors' ").lower()
    if user == 'quit':
        print("Game ended :(")
        exit()
    computer = random.choice(['rock', 'paper','scissors'])
    print(computer)
    if user == computer:
        return 'tie'
    elif is_win(user, computer):
        return 'You win!'
    else:
        return 'You lost!'

def is_win(player,opponent):
    if (player == 'rock'and opponent == 'scissors') or (player == 'scissors'and opponent == 'paper') or \
            (player=='paper' and opponent == 'rock'):
        return True
    return False

while True:
    if start_game != 'start':
        start_game = input('Sorry i dont understand please try again:')
    else:
        print(play())

尝试进行上述更改，看看它是否适合您。

【讨论】：

非常感谢！！我只是想知道为什么我在 return False 之前不需要 else: 语句？
@Harini 因为如果'if'语句为真，那么函数将返回并且函数的其余部分将不会被执行。所以有一个“其他”不会有任何区别

【解决方案2】：

你在定义 is_win() 函数之前使用它！

在播放函数之前定义函数！

考虑一下

start_game = input('Type start to start game:').lower()
import random

def is_win(player,opponent):
    if (player == 'rock'and opponent == 'scissors') or (player == 'scissors'and opponent == 'paper') or \
            (player=='paper' and opponent == 'rock'):
        return True
    print(play())

def play():
    user = input("'rock' , 'paper', 'scissors' ").lower()
    if user == 'quit':
        print("Game ended :(")
        exit()
    computer = random.choice(['rock', 'paper','scissors'])
    print(computer)
    if user == computer:
        return 'tie'
    elif not user == is_win(user, computer):
        return 'You win!'
    else:
        return 'You lost!'


while True:
    if start_game != 'start':
        start_game = input('Sorry i dont understand please try again:')
    else:
        print(play())

【讨论】：

这不是问题。他们在另一个函数中调用函数“is_win”。这两个函数都已定义，并且可以在编译器到达“while True”时使用。

【解决方案3】：

print(play()) 是干什么用的？您正在未完成的一轮中运行另一轮游戏。我认为您应该返回 False 而不是那个，并更改“elif”条件：

start_game = input('Type start to start game:').lower()
import random
def play():
    user = input("'rock' , 'paper', 'scissors' ").lower()
    if user == 'quit':
        print("Game ended :(")
        exit()
    computer = random.choice(['rock', 'paper','scissors'])
    print(computer)
    if user == computer:
        return 'tie'
    elif is_win(user, computer):
        return 'You win!'
    else:
        return 'You lost!'

def is_win(player,opponent):
    if (player == 'rock'and opponent == 'scissors') or (player == 'scissors'and opponent == 'paper') or \
            (player=='paper' and opponent == 'rock'):
        return True
    return False

while True:
    if start_game != 'start':
        start_game = input('Sorry i dont understand please try again:')
    else:
        print(play())

【讨论】：