Paul Rooney解决： 在xcode的顶部放置了
的全局我已经搜索了其他网站和整个堆栈器溢出并且无法找到解决方案，这是一个错误的独特案例，因为我无法在定义中声明变量。 每当我开始我的游戏时，让我们说例如我输入1,4作为网格参考，它将回复
“UnboundLocalError：'在赋值前引用'的局部变量'
我该如何解决这个问题？
代码：
import time
while 1==1:
cont=1
while cont==1:
of="-"
tf="-"
rf="-"
ov="-"
tv="-"
rv="-"
os="-"
ts="-"
rs="-"
go=1
def grid(g):
print (" ")
print (" 1 2 3")
print ("4 {} | {} | {}". format(of,tf,rf))
print (" --------")
print ("5 {} | {} | {}". format(ov,tv,rv))
print (" --------")
print ("6 {} | {} | {}". format(os,ts,rs))
print (" ")
def xturn(x):
go=1
while go==1:
goo=input("X-Insert Board Reference: ")
go=0
if goo in "1,4" and of not in "o":
of="x"
elif goo in "2,4" and tf not in "o":
tf="x"
elif goo in "3,4" and rf not in "o":
rf="x"
elif goo in "1,5" and ov not in "o":
ov="x"
elif goo in "2,5" and tv not in "o":
tv="x"
elif goo in "3,5" and rv not in "o":
rv="x"
elif goo in "1,6" and os not in "o":
os="x"
elif goo in "2,6" and ts not in "o":
ts="x"
elif goo in "3,6" and rs not in "o":
rs="x"
else:
print ("NOT RECOGNISED/POSITION TAKEN")
go=1
def oturn(o):
go=1
while go==1:
goo=input("O-Insert Board Reference: ")
go=0
if goo in "1,4" and of not in "x":
of="o"
elif goo in "2,4" and tf not in "x":
tf="o"
elif goo in "3,4" and rf not in "x":
rf="o"
elif goo in "1,5" and ov not in "x":
ov="o"
elif goo in "2,5" and tv not in "x":
tv="o"
elif goo in "3,5" and rv not in "x":
rv="o"
elif goo in "1,6" and os not in "x":
os="o"
elif goo in "2,6" and ts not in "x":
ts="o"
elif goo in "3,6" and rs not in "x":
rs="o"
else:
print ("NOT RECOGNISED/POSITION TAKEN")
go=1
def xcheck(xc):
if of in "x" and tf in "x" and rf in "x":
print ("X WINS!")
cont=0
elif ov in "x" and tv in "x" and rv in "x":
print ("X WINS!")
cont=0
elif os in "x" and ts in "x" and rs in "x":
print ("X WINS!")
cont=0
elif of in "x" and ov in "x" and os in "x":
print ("X WINS!")
cont=0
elif tf in "x" and tv in "x" and ts in "x":
print ("X WINS!")
cont=0
elif rf in "x" and rv in "x" and rs in "x":
print ("X WINS!")
cont=0
elif of in "x" and tv in "x" and rs in "x":
print ("X WINS!")
cont=0
elif rf in "x" and tv in "x" and os in "x":
print ("X WINS!")
cont=0
else:
grid(1)
oturn(1)
def ocheck(oc):
if of in "o" and tf in "o" and rf in "o":
print ("O WINS!")
cont=0
elif ov in "o" and tv in "o" and rv in "o":
print ("O WINS!")
cont=0
elif os in "o" and ts in "o" and rs in "o":
print ("O WINS!")
cont=0
elif of in "o" and ov in "o" and os in "o":
print ("O WINS!")
cont=0
elif tf in "o" and tv in "o" and ts in "o":
print ("O WINS!")
cont=0
elif rf in "o" and rv in "o" and rs in "o":
print ("O WINS!")
cont=0
elif of in "o" and tv in "o" and rs in "o":
print ("O WINS!")
cont=0
elif rf in "o" and tv in "o" and os in "o":
print ("O WINS!")
cont=0
else:
grid(1)
xturn(1)
ocheck(1)
xcheck(1)
ocheck(1)
xcheck(1)
ocheck(1)
xcheck(1)
ocheck(1)
xcheck(1)
ocheck(1)
xcheck(1)
print ("THERE WAS NO WINNER")
回溯：
Traceback (most recent call last):
File "b4assign.py", line 141, in <module>
ocheck(1)
File "b4assign.py", line 138, in ocheck
xturn(1)
File "b4assign.py", line 33, in xturn
if goo in "1,4" and of not in "o":
UnboundLocalError: local variable 'of' referenced before assignment
答案 0 :(得分：2)
即使仅在外部范围内定义变量，访问其中一个函数内的变量也基本可行。但是在xturn函数中，您有一行分配给of。在发生错误之前是否已执行此行并不重要;它仅存在于函数内的任何地方都会导致Python解释器将其视为局部变量。因此，当在if子句中访问它时，Python会尝试访问本地变量of，现在确实在此之前尚未分配这样的局部变量。
答案 1 :(得分：1)
的变量（以及所有其他双字母变量）在 xturn 功能中不可用。
我强烈建议您使用增量编程：编写几行代码，使其工作，然后扩展您的程序。这使您可以非常轻松地将错误归零。每当我攻击一种新的编程风格时，我发现这非常有用。您似乎对此处使用的某些技术不熟悉。
例如，您使用 而不是 == 进行比较。作为一般原则，这不会很好。
在主程序之前声明你的函数。你写这篇文章的方式，每次循环都重新定义你的函数。将函数移动到顶部也可以解决许多变量范围问题。
学习使用布尔值和变量。你的循环应该是这样的：
while True:
cont = True
while cont:
通过将变量作为参数传递给函数，可以使变量可用。我可以看到你对此不熟悉，因为你已经为这个函数提供了一个你永远不会使用的参数 x 。
总的来说，你不应该有9个变量：你应该有一个列表，然后将整个列表作为游戏板的当前状态传递。如果您将方块编号为0-8，则可以在这方面轻松使用该板。
要解决当前问题，您可以将此行添加到您的每个例程中：
global of,tf,rf,ov,tv,rv,os,ts,rs
这将使变量可用。我看到@Thomas已经指出了这一点。
不过，我鼓励你更干净地设计这个。使用全局变量通常是糟糕的设计。另外，请注意您需要为此程序复制多少代码？它应该更容易。
答案 2 :(得分：0)
在函数xturn中，使用在该范围内未声明/已知的变量of，从而导致错误。
如果查看堆栈跟踪：
Traceback (most recent call last):
File "D:/Projekte/Python/snakes35/blabla.py", line 141, in <module>
ocheck(1)
File "D:/Projekte/Python/snakes35/blabla.py", line 138, in ocheck
xturn(1)
File "D:/Projekte/Python/snakes35/blabla.py", line 33, in xturn
if goo in "1,4" and of not in "o":
UnboundLocalError: local variable 'of' referenced before assignment
你可以通过查看文件中出现错误的行来解决这个问题。