Question

我正在尝试使第一个功能opt1（）使健康降低50点，但是由于某种原因它不起作用。代码不在函数中时起作用。无效的行：health = int(health) - 50

import random


health = "100"
p = "1"


def opt1():
   print("Du förlorade striden")
   health = int(health) - 50
   print(health)


def opt2():
   print("hej2")


def opt3():
   print("hej3")


q = [opt1, opt2, opt3]
ind = random.randint(0,len(q)-1)


print("you have ", p, " potions")
print("Your health is ", health,)
while True:
   print("Commands: Add, drink, sleep, fight, quest")
   a = input("Enter a command: ")
   if a == "add":
       health = int(health)
       p = int(p) + 1
       print("you have ", p, " potions")
       print("Your health is ", health,)
   if a == "fight":
       q[ind]()

Answer 1

因此我将health变量更改为一个整数。

并将global添加到函数中：

import random

health = 100
p = "1"


def opt1():
    global health
    print("Du förlorade striden")
    health -= 50
    print(health)


def opt2():
    print("hej2")


def opt3():
    print("hej3")


q = [opt1, opt2, opt3]
ind = random.randint(0,len(q)-1)


print("you have ", p, " potions")
print("Your health is ", health,)
while True:
    print("Commands: Add, drink, sleep, fight, quest")
    a = input("Enter a command: ")
    if a == "add":
        health = int(health)
        p = int(p) + 1
        print("you have ", p, " potions")
        print("Your health is ", health,)
    if a == "fight":
        q[ind]()

此外，您可以在Python here中的scoping上阅读更多内容。

希望这对您有帮助！

Answer 2

def opt1():

在opt1中，您应该声明要提供给函数的变量。

例如：

def opt1(health):

函数中的积分变量不起作用

2 个答案: