我一般对乌龟和编程都不熟悉,所以想制作一个标签游戏。有两个用户,一个是奔跑者,一个是追逐者(鲍勃和菲尔)。当追赶者靠近跑步者(使用距离公式检测到)时,程序会将其重新设置为原始开始位置。但是,每当我运行该程序时,Turtle Graphics窗口就会冻结并停止响应,最终导致我关闭该程序。谁能帮我找出问题所在?
import turtle
import math
bob = turtle.Turtle()
wn = turtle.Screen()
wn.bgcolor("black")
wn.setup(width = 700, height = 700)
wn.tracer()
border_pen = turtle.Turtle()
border_pen.speed(0)
border_pen.color("white")
border_pen.penup()
border_pen.setposition(-300,-300)
border_pen.pendown()
border_pen.pensize(3)
for side in range(4):
border_pen.fd(600)
border_pen.lt(90)
border_pen.hideturtle()
bob.shape("triangle")
bob.speed(0)
bob.color("blue")
bob.penup()
bob.setpos(-50, 0)
def draw():
bob.penup()
def nodraw():
bob.pendown()
def fd():
bob.fd(20)
if bob.xcor() > 280:
bob.setx(280)
if bob.ycor() > 280:
bob.sety(280)
if bob.xcor() < -280:
bob.setx(-280)
if bob.ycor() <- 280:
bob.sety(-280)
def right():
bob.right(90)
def left():
bob.left(90)
turtle.listen()
turtle.onkey(left, "Left")
turtle.onkey(fd, "Up")
turtle.onkey(right, "Right")
phil = turtle.Turtle()
phil.speed(0)
phil.penup()
phil.shape("triangle")
phil.color("red")
phil.setpos(50, 0)
def fdp():
phil.fd(20)
if phil.xcor() > 280:
phil.setx(280)
if phil.ycor() > 280:
phil.sety(280)
if phil.xcor() < -280:
phil.setx(-280)
if phil.ycor() <- 280:
phil.sety(-280)
def rightp():
phil.right(90)
def leftp():
phil.left(90)
turtle.listen()
turtle.onkey(leftp, "a")
turtle.onkey(fdp, "w")
turtle.onkey(rightp, "d")
def isCollision(t1, t2):
distance = math.sqrt(math.pow(t1.xcor()-t2.xcor(),2)+math.pow(t1.ycor()-t2.ycor(),2))
if distance < 20:
return True
else:
return False
while True:
if isCollision(bob, phil):
bob.setposition(-50, 0)
phil.setposition(50, 0)
turtle.done()
答案 0 :(得分:0)
经过审查,这应该可行:
import turtle
import math
bob = turtle.Turtle()
wn = turtle.Screen()
wn.bgcolor("black")
wn.setup(width = 700, height = 700)
wn.tracer()
border_pen = turtle.Turtle()
border_pen.speed(0)
border_pen.color("white")
border_pen.penup()
border_pen.setposition(-300,-300)
border_pen.pendown()
border_pen.pensize(3)
for side in range(4):
border_pen.fd(600)
border_pen.lt(90)
border_pen.hideturtle()
bob.shape("triangle")
bob.speed(0)
bob.color("blue")
bob.penup()
bob.setpos(-50, 0)
def draw():
bob.penup()
def nodraw():
bob.pendown()
def fd():
bob.fd(20)
if bob.xcor() > 280:
bob.setx(280)
if bob.ycor() > 280:
bob.sety(280)
if bob.xcor() < -280:
bob.setx(-280)
if bob.ycor() <- 280:
bob.sety(-280)
distance = math.sqrt(math.pow(bob.xcor()-phil.xcor(),2)+math.pow(bob.ycor()-phil.ycor(),2))
print(distance)
if distance < 20:
bob.setposition(-50, 0)
phil.setposition(50, 0)
def right():
bob.right(90)
def left():
bob.left(90)
turtle.listen()
turtle.onkey(left, "Left")
turtle.onkey(fd, "Up")
turtle.onkey(right, "Right")
phil = turtle.Turtle()
phil.speed(0)
phil.penup()
phil.shape("triangle")
phil.color("red")
phil.setpos(50, 0)
def fdp():
phil.fd(20)
if phil.xcor() > 280:
phil.setx(280)
if phil.ycor() > 280:
phil.sety(280)
if phil.xcor() < -280:
phil.setx(-280)
if phil.ycor() <- 280:
phil.sety(-280)
def rightp():
phil.right(90)
def leftp():
phil.left(90)
turtle.listen()
turtle.onkey(leftp, "a")
turtle.onkey(fdp, "w")
turtle.onkey(rightp, "d")
turtle.done()
答案 1 :(得分:0)
使您的turtle.done()
缩进while
循环中。这将解决您的冻结问题。像这样:
while True:
if isCollision(bob, phil):
bob.setposition(-50, 0)
phil.setposition(50, 0)
turtle.done()
您也可以删除while
循环,因为您实际上不需要它来使if
语句起作用。像这样:
if isCollision(bob, phil):
bob.setposition(-50, 0)
phil.setposition(50, 0)
turtle.done()
答案 2 :(得分:0)
您的原始代码和我在下面尝试解决的建议答案存在问题。具体来说:您无需计算乌龟之间的距离,因为它们已经知道该怎么做;您和其他人使用if (message.author.bot) return;
没有意义,因为您没有tracer()
的电话;鲍勃和菲尔之间的碰撞应该是对称的,但答案之一是鲍勃与菲尔相撞,而不是菲尔与鲍勃相撞。答案之一似乎并不了解update()
的本质。
turtle.done()