pygame中的自制二维碰撞检测不起作用

时间:2018-09-09 19:11:26

标签: python python-3.x list pygame

我正在制作一个类似tron的游戏,两个玩家使用玩家1的按键w,a,s,d控制他们的角色,而玩家2则使用上,下,左,右来控制他们的角色。在玩家移动时,用户进行追踪,并且如果对方玩家触摸玩家的踪迹,则表示他们“不在”。

游戏代码:

from steel2D import *

window = Window(600, 450, "Steel 2D Tutorial")
running = True
char1 = Rect(window, 0, 0, 5, 5, (255, 0, 0))
char2 = Rect(window, 595, 445, 5, 5, (0, 0, 255))
trail1 = []
trail2 = []
red = 0
blue = 0

while running:
    sleep(2)
    keys = OnKeyInput()

    if keys[pygame.K_w]:
        char1.y -= 2.2
        trail1.append(char1)
    if keys[pygame.K_a]:
        char1.x -= 2.2
        trail1.append(char1)
    if keys[pygame.K_s]:
        char1.y += 2.2
        trail1.append(char1)
    if keys[pygame.K_d]:
        char1.x += 2.2
        trail1.append(char1)

    if keys[pygame.K_UP]:
        char2.y -= 2.2
        trail2.append(char2)
    if keys[pygame.K_LEFT]:
        char2.x -= 2.2
        trail2.append(char2)
    if keys[pygame.K_DOWN]:
        char2.y += 2.2
        trail2.append(char2)
    if keys[pygame.K_RIGHT]:
        char2.x += 2.2
        trail2.append(char2)

    if char1.x >= 595:
        char1.x = 595
    if char1.x <= 0:
        char1.x = 0
    if char1.y >= 445:
        char1.y = 445
    if char1.y <= 0:
        char1.y = 0

    if char2.x >= 595:
        char2.x = 595
    if char2.x <= 0:
        char2.x = 0
    if char2.y >= 445:
        char2.y = 445
    if char2.y <= 0:
        char2.y = 0

    char1.draw()
    char2.draw()
    Update()

    for t1, t2 in zip(trail1, trail2):
        if OnRectCollison(t1, t2):
            exit()

    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            running = False
            pygame.quit()

在每次按键时,用户跟踪都会添加到列表中,然后循环遍历以检查是否存在冲突。现在看来,只有在玩家的第一个足迹碰到时,碰撞才会发生。

steel2D(我的游戏引擎)的代码:

import pygame
from threading import Thread

pygame.init()


def sleep(ms):
    pygame.time.delay(ms)


def Window(w, h, title="steel2D"):
    window = pygame.display.set_mode((w, h))
    pygame.display.set_caption(title)
    return window


def Update():
    pygame.display.update()


def OnKeyInput():
    return pygame.key.get_pressed()


class Rect:
    def __init__(self, window, x, y, w, h, color=(0, 0, 0)):
        self.window = window
        self.x = x
        self.y = y
        self.w = w
        self.h = h
        self.color = color

    def draw(self):
        pygame.draw.rect(self.window, self.color, (self.x, self.y, self.w, self.h))


class Circle:
    def __init__(self, window, x, y, r, color=(0, 0, 0)):
        self.window = window
        self.x = x
        self.y = y
        self.r = r
        self.color = color

    def draw(self):
        pygame.draw.circle(self.window, self.color, (self.x, self.y), self.r, self.r)


def Line(window, color, sx, sy, ex, ey, width=2):
    pygame.draw.line(window, color, (sx, sy), (ex, ey), width)


def OnRectCollison(rect1, rect2):
    if rect1.x < rect2.x + rect2.w and rect1.x + rect1.w > rect2.x and rect1.y < rect2.y + rect2.h and rect1.y + rect1.h > rect2.y:
        return True
    else:
        return False

任何帮助将不胜感激。

1 个答案:

答案 0 :(得分:1)

有几个问题:

    如果要比较头矩形与trail1trail2中的每个矩形,则
  1. zip是错误的功能。您只会在相同的索引处获得成对的rect,并且还会删除较长列表中的元素。看一下这个例子:

    >>> trail1 = [1, 2, 3]
    >>> trail2 = ['a', 'b', 'c', 'd', 'e']
    >>> list(zip(trail1, trail2))
    [(1, 'a'), (2, 'b'), (3, 'c')]
    

    只需遍历路径列表中的矩形并检查char1char2是否与它们冲突。

    # I skip the last few rects by slicing the list, otherwise they would collide immediately.
    for trail in trail1[:-5]:
        if OnRectCollison(char1, trail):
            print('collision')
    for trail in trail2:
        if OnRectCollison(char1, trail):
            print('collision')
    
  2. 在移动时,会将相同的Rect对象(char1char2)添加到跟踪列表,因此列表仅包含对头部的引用,因此仅步道的头部能够碰撞。

    trail1.append(char1)
    

    添加(浅)副本:

    trail1.append(copy.copy(char1))
    

侧面说明:碰撞的拼写错误。