当我的蛇碰到我时，苹果不会消失

Question

这是我的第一篇帖子！我想在Pygame中复制一条蛇，但是当我的蛇碰到苹果时，它就不能传送到另一个地方。这是我的密码：

import pygame          
import time
import random

pygame.init()          

wh = (255,255,255)     #white
bk = (0,0,0)           #black
ylw = (255,255,0)      #yellow 
prpl = (127,0,127)     #purple

snakeXChange = 0
snakeYChange = 0
dWidth = 800
dHeight = 600 
clock = pygame.time.Clock()
fps = 25
blockSize = 15
appleX = round(random.randrange(0, dWidth - blockSize)/ 10.0)* 10.0
appleY = round(random.randrange(0, dHeight - blockSize) / 10.0)* 10.0
snakeX = dWidth / 2           
snakeY = dHeight / 2          
font = pygame.font.SysFont(None, 25)
def message(msg, color):
    screenText = font.render(msg, True, color)
    gameDisplay.blit(screenText, [dWidth / 2, dHeight / 2])

gameDisplay = pygame.display.set_mode((dWidth,dHeight))     
pygame.display.set_caption('Snek')                   #shoutout to Hopson


gameExit = False                                    
gameOver = False


while not gameExit:                                  
    for event in pygame.event.get():                 
        if event.type == pygame.QUIT:                
            gameExit = True                         
        if event.type == pygame.KEYDOWN:
            if event.key == pygame.K_LEFT:
                snakeXChange = -blockSize
                snakeYChange = 0
            if event.key == pygame.K_RIGHT:
                snakeXChange = blockSize
                snakeYChange = 0
            if event.key == pygame.K_UP:
                snakeYChange = -blockSize
                snakeXChange = 0
            if event.key == pygame.K_DOWN:
                snakeYChange = blockSize
                snakeXChange = 0


    snakeX += snakeXChange

    snakeY += snakeYChange

    gameDisplay.fill(prpl)                          

    pygame.draw.rect(gameDisplay, wh, [snakeX,snakeY,blockSize,blockSize])  #snake

    pygame.draw.rect(gameDisplay, ylw, [appleX,appleY,blockSize,blockSize]) #apple

    if snakeX == appleX and snakeY == appleY:
        appleX = round(random.randrange(0, dWidth - blockSize)/ 10.0)* 10.0
        appleY = round(random.randrange(0, dHeight - blockSize) / 10.0)* 10.0

    clock.tick(fps)

    pygame.display.update()  

message("GAME OVER NIBBA", ylw)
pygame.display.update()                              #updates the display
time.sleep(5)
pygame.quit()                                        
quit()    

原始来源：https://paste.ee/p/eSimt

我遵循newboston的教程作为参考。你知道吗

干杯

萨尔蒙桑

Answer 1

你将appleX和appleY坐标四舍五入到10的倍数，但是以15像素的步长移动，这意味着蛇可能无法触摸苹果。例如，appleX可能是410，蛇从400开始，所以它永远无法到达苹果。你知道吗

将所有位置四舍五入到15的最接近倍数（即blockSize和速度）：

appleX = round(random.randrange(0, dWidth - blockSize) / 15.0) * 15.0
appleY = round(random.randrange(0, dHeight - blockSize) / 15.0) * 15.0
snakeX = round(dWidth / 2. / 15.) * 15
snakeY = round(dHeight / 2. / 15.) * 15

也可以将15作为step参数传递给random.randrange，得到15的随机倍数。你知道吗

appleX = random.randrange(0, dWidth-blockSize, 15)

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也可以使用^{}s进行碰撞检测并存储坐标。你知道吗