在python中永远不会到达 Else

Question

我只需要这样做:

1. 接受用户输入的字符串，并像上一阶段一样打印游戏网格。

2. 分析游戏状态并打印结果

:

1. Enter cells: XXXOO__O_    ---------  
| X X X |  
| O O _ |  
| _ O _ |    ---------
X wins

问题是我的程序从来没有达到elif语句，我不知道为什么。

n = input()
number_grid = []
Winner = ""
print("---------")
Cont = {"X": 0, "O": 0, "_": 0}   # This will count to check for impossible games
for i in range(0, len(n), 3):
print("|", n[i], n[i+1], n[i+2], "|")
number_grid.append([n[i], n[i+1], n[i+2]])
Cont[n[i]] = Cont[n[i]] + 1
Cont[n[i + 1]] = Cont[n[i + 1]] + 1
Cont[n[i + 2]] = Cont[n[i + 2]] + 1
print("---------")
Impossible = False
if 2 <= Cont["X"] - Cont["O"] or Cont["X"] - Cont["O"] <= -2:
Impossible = True
print("Impossible")
exit()
if Winner == "" and Impossible is False:
for rows in range(0, 3):    # Check winner in rows
if number_grid[rows][0] == number_grid[rows][1] == number_grid[rows][2] != "_":
Winner = number_grid[rows][0]
print(Winner, "wins")
for columns in range(0, 3):  # Check winner in columns
if number_grid[0][columns] == number_grid[1][columns] == number_grid[2][columns] != "_":
Winner = number_grid[0][columns]
print(Winner, "wins")
if number_grid[0][0] == number_grid[1][1] == number_grid[2][2] != "_":  # Check winner in first diagonal
Winner = number_grid[0][0]
print(Winner, "wins")
if number_grid[0][2] == number_grid[1][1] == number_grid[2][0] != "_":  # Check winner in second diagonal
Winner = number_grid[0][2]
print(Winner, "wins")
elif Cont["_"] != 0:
print("Game not finished")
elif Cont["X"] + Cont["O"] == 9:
print("Draw")

Answer 1

减少你的代码…

Winner = ""
Impossible = False
if Winner == "" and Impossible is False:
# Stuff
elif Cont["_"] != 0:
print("Game not finished")
elif Cont["X"] + Cont["O"] == 9:
print("Draw")

在您当前的代码中，Winner != ""或Impossible != True是不可能的。因此，您永远不会输入elif语句。