Python海龟碰撞检测通过；不在「；列表为什么'；这不管用吗？

此测试代码的目标是使玩家能够使用W、a、S、D移动，并使用ENTER键开始或停止构建墙。如果有人能告诉我为什么他只是偶尔撞到墙壁，我将不胜感激。在一般意义上也可以随意评论我的代码！提前谢谢

import turtle

grid_size = 10

t1 = turtle.Pen()
t1.width(grid_size)
t1.up()

walls = [[0,0]]
walls.clear()

def toggle_building():
    if t1.isdown():
        t1.up()
    else:
        t1.down()

def lay_brick():
    if t1.isdown() and t1.pos() not in walls:
        walls.append(t1.pos())
    print("Brick layed.")

def print_pos():
    print(t1.pos())

def move_up():
    t1.setheading(90)
    if t1.pos() + [0, grid_size] not in walls:
        t1.forward(grid_size)
        lay_brick()
    else:
        print("wall")
    print_pos()
def move_left():
    t1.setheading(180)
    if t1.pos() - [grid_size, 0] not in walls:
        t1.forward(grid_size)
        lay_brick()
    else:
        print("wall")
    print_pos()
def move_down():
    t1.setheading(270)
    if t1.pos() - [0, grid_size] not in walls:
        t1.forward(grid_size)
        lay_brick()
    else:
        print("wall")
    print_pos()
def move_right():
    t1.setheading(0)
    if t1.pos() + [grid_size, 0] not in walls:
        t1.forward(grid_size)
        lay_brick()
    else:
        print("wall")
    print_pos()

turtle.onkeypress(move_up, "w")
turtle.onkeypress(move_left, "a")
turtle.onkeypress(move_down, "s")
turtle.onkeypress(move_right, "d")
turtle.onkeypress(toggle_building, "Return")
turtle.listen()

---

正如在你的问题的评论中所提到的，海龟在浮点平面上游荡，即使你认为它已经回到了与以前完全相同的位置，但始终会有一点浮点噪声添加到该位置。将位置转换为

int

进行比较肯定有帮助，但可能还不够

下面是我的尝试，通过改变坐标系本身，使其变得更加健壮，这样看起来我们移动的是1，而不是10。它还试图减少浮点到整数转换所需的位数：

from turtle import Screen, Turtle

GRID_SIZE = 10

def toggle_building():
    if turtle.isdown():
        turtle.penup()
    else:
        turtle.pendown()

def lay_brick(position):
    if turtle.isdown():
        walls.add(position)

def move_up():
    turtle.setheading(90)

    position = int(turtle.xcor()), int(turtle.ycor()) + 1

    if position not in walls:
        turtle.goto(position)
        lay_brick(position)

def move_left():
    turtle.setheading(180)

    position = int(turtle.xcor()) - 1, int(turtle.ycor())

    if position not in walls:
        turtle.goto(position)
        lay_brick(position)

def move_down():
    turtle.setheading(270)

    position = int(turtle.xcor()), int(turtle.ycor()) - 1

    if position not in walls:
        turtle.goto(position)
        lay_brick(position)

def move_right():
    turtle.setheading(0)

    position = int(turtle.xcor()) + 1, int(turtle.ycor())

    if position not in walls:
        turtle.goto(position)
        lay_brick(position)

screen = Screen()
WIDTH, HEIGHT = (screen.window_width() / 2) // GRID_SIZE, (screen.window_height() / 2) // GRID_SIZE
screen.setworldcoordinates(-WIDTH, -HEIGHT, WIDTH, HEIGHT)

turtle = Turtle()
turtle.width(GRID_SIZE)
turtle.fillcolor('red')
turtle.penup()

walls = set()

screen.onkeypress(move_up, "w")
screen.onkeypress(move_left, "a")
screen.onkeypress(move_down, "s")
screen.onkeypress(move_right, "d")
screen.onkeypress(toggle_building, "Return")

screen.listen()

screen.mainloop()

---

它还使用

集

来包含墙。由于顺序无关紧要，这将使测试更快。

是不是

pos（）

是一个浮点数？如果我没有弄错的话，pos（）返回一个带有两个浮点数的Vec2D，如下所示：[20.00，10.00]@RobertHarveyFloating point number无法使用equals进行可靠比较。诚然，如果您直接存储

pos（）

，这并不重要，但您正在那里做一些算术，所以。谢谢@RobertHarvey！我添加了一些代码将浮点转换为整数，然后将它们附加到列表中，再添加一些代码以确保整数始终是10的倍数。