Question

def user(choose):
    if (choose == "1"):
        play = game()
    elif (choose == "2"):
        return stats(play)
    else:
        return quit()

我想从game（）函数中获取值并在stats（）中使用它，但是我收到一条错误消息，说未定义游戏。如何声明game（）并在其他函数中使用它？

Answer 1

您可以“推迟”执行func1：

def func1():
    return 'abc'

def something_else(callable):
    callable()

def main():
    hello = None

    def f():
        """set result of func1 to variable hello"""
        nonlocal hello
        hello = func1()

    # over here func1 is not executed yet, hello have not got its value
    # you could pass function f to some other code and when it is executed, 
    # it would set result for hello 
    print(str(hello))  # would print "None"
    call_something_else(f)
    print(str(hello))  # would print "abc"

main()

问题改变之后...

现在，您的本地变量play超出了统计范围。此外，看起来您希望该函数将被调用两次。您需要在全局内容中保存play

play = None  # let's set it to some default value
def user(choose):
    global play  # let python know, that it is not local variable

    if choose == "1":  # no need for extra brackets
        play = game()

    if choose == "2" and play:  # double check that play is set
        return stats(play)

    return quit()

声明变量以使其运行而不运行

1 个答案: