我试图让这个石头剪刀布游戏返回一个布尔值,如 set 中所示player_wins
True 或 False,具体取决于玩家是否获胜,或者完全重构此代码以使其不使用 while 循环。
我来自世界的系统管理员,所以如果这是以错误的风格编写的,请保持温和。
我已经尝试了一些方法,并且我了解 TIMTOWTDI,并且只需要一些输入。
Thanks.
import random
global player_wins
player_wins=None
def rps():
player_score = 0
cpu_score = 0
while player_score < 3 and cpu_score < 3:
WEAPONS = 'Rock', 'Paper', 'Scissors'
for i in range(0, 3):
print "%d %s" % (i + 1, WEAPONS[i])
player = int(input ("Choose from 1-3: ")) - 1
cpu = random.choice(range(0, 3))
print "%s vs %s" % (WEAPONS[player], WEAPONS[cpu])
if cpu != player:
if (player - cpu) % 3 < (cpu - player) % 3:
player_score += 1
print "Player wins %d games\n" % player_score
else:
cpu_score += 1
print "CPU wins %d games\n" % cpu_score
else:
print "tie!\n"
rps()
我正在尝试做这样的事情:
print "%s vs %s" % (WEAPONS[player], WEAPONS[cpu])
if cpu != player:
if (player - cpu) % 3 < (cpu - player) % 3:
player_score += 1
print "Player wins %d games\n" % player_score
if player_score == 3:
return player_wins==True
else:
cpu_score += 1
print "CPU wins %d games\n" % cpu_score
if cpu_score == 3:
return player_wins==False
else:
print "tie!\n"
忽略重构问题,您需要了解函数和返回值。你根本不需要全局。曾经。你可以这样做:
def rps():
# Code to determine if player wins
if player_wins:
return True
return False
然后,只需为该函数外部的变量赋值,如下所示:
player_wins = rps()
它将被分配您刚刚调用的函数的返回值(True 或 False)。
在评论之后,我决定用惯用的方式添加这一点,这样表达会更好:
def rps():
# Code to determine if player wins, assigning a boolean value (True or False)
# to the variable player_wins.
return player_wins
pw = rps()
这分配了布尔值player_wins
(函数内部)到pw
函数外的变量。
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