python - 让 turtle 保持在一个正方形内，没有输出，也没有错误

Question

我希望 turtle 不要出格

窗口出现并且没有错误，但加载仍然存在，如果我单击该窗口，它就会关闭。

• x, y 是 turtle 的位置。
• D 是显示天气或 turtle 是否在方 block 内且在线(？)。

• a 没什么

  1. 方向改变
  2. 速度变化
  3. 位置检查和更改
  4. 如果 turtle 在一条线上，它就会沿着边界移动。

import turtle
import msvcrt

turtle.setup(100, 100, 0, 0)
t = turtle.Turtle()
D = 0
a = 1

while True:
    if msvcrt.kbhit():
        c = msvcrt.getch().decode('UTF - 8')
        if c == 'j': # turn left
            t.left(10)
        if c == 'k':
            t.right(10)
        if c == 'a':# speed change
            a += 1
            if a > 10: # speed limitation
                a = 10
        if c == 'z':
            a += -1
            if a < 0:
                a = 0
    #---------------------------------------
    x = t.xcor() # turtle's location
    y = t.ycor()
    if x > 100: # if go out of square, go back to the square
        t.setx(100)
    if x < -100:
        t.setx(-100)
    if y > 100:
        t.sety(100)
    if y < -100:
        t.sety(-100)
    #---------------------------------------
    x = t.xcor() # turtle's location
    y = t.ycor()
    h = t.heading() # turtle's direction
    #---------------------------------------
    if - 100 < x < 100 and -100 < y < 100:
        D = 0 # if it's in the square, D = 0
    elif x == 100 and -100 < y < 100: # if it's in ~
        if 0 <= d <= 90: # direction changes
            t.setheading(90) # direction change to 90 degrees
            D = 1 # D = 1
        elif 270 <= d < 360:
            t.setheading(270)
            D = 2
    elif x == -100 and -100 < y < 100:
        if 90 <= d < 180:
            t.setheading(90)
            D = 2
        elif 180 <= d <= 270:
            t.setheading(270)
            D = 1
    elif y == 100 and -100 < x < 100:
        if 0 <= d < 90:
            t.setheading(0)
            D = 2
        elif 90 <= d <= 180:
            t.setheading(180)
            D = 1
    elif y == -100 and -100 < x < 100:
        if 180 <= d < 270:
            t.setheading(180)
            D = 2
        elif 270 <= d < 360:
            t.setheading(0)
            D = 1
    elif D == 1:
        t.left(90)
    elif D == 2:
        t.right(90)

Answer 1

这段代码一团糟。首先，通过将窗口大小设置为 100 x 100，然后在两个维度上在 -100 和 100 之间移动 turtle ，只有 1/4 的游戏区域可见!为什么使用 msvcrt当turtle内置了完美的键盘事件时模块？据我所知，您的代码检查并纠正了 turtle 的位置，但从未真正移动它!你有一个 while True:循环在像 turtle 这样的事件驱动环境中没有地位。

让我们从一个可以通过键盘操纵的正方形内运动的简单示例开始:

from turtle import Screen, Turtle

def turn_left():
    turtle.left(10)

def turn_right():
    turtle.right(10)

def speed_up():
    speed = turtle.speed() + 1

    if speed > 10: # speed limitation
        speed = 10

    turtle.speed(speed)

def slow_down():
    speed = turtle.speed() - 1

    if speed < 1: # speed limitation
        speed = 1

    turtle.speed(speed)

def move():
    # there are two different senses of 'speed' in play, we'll exploit
    #  both!  I.e. as we move faster, we'll draw faster and vice versa
    turtle.forward(turtle.speed())

    # if we go out of square, go back into the square
    if not -100 < turtle.xcor() < 100:
        turtle.undo()
        turtle.setheading(180 - turtle.heading())
    elif not -100 < turtle.ycor() < 100:
        turtle.undo()
        turtle.setheading(360 - turtle.heading())

    screen.ontimer(move, 100)

screen = Screen()
screen.setup(300, 300)

# show turtle's boundary
boundary = Turtle('square', visible=False)
boundary.color('pink')
boundary.shapesize(10)
boundary.stamp()

turtle = Turtle()

move()

screen.onkey(turn_left, 'j')
screen.onkey(turn_right, 'k')
screen.onkey(speed_up, 'a')
screen.onkey(slow_down, 'z')

screen.listen()
screen.mainloop()