当我的蛇触摸它时，苹果不会消失

Question

这是我的第一篇文章！我试图在Pygame中制作蛇克隆，但是当我的蛇触摸苹果时，它不会传送到另一个地方。这是我的代码：

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

我正在遵循NewBoston的教程作为参考。

干杯

鲑鱼桑

Answer 1

您将appleX和appleY坐标四舍五入到最接近的 10 的倍数，但以 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。

只需将所有位置四舍五入到最接近的 15 的倍数(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

您也可以将15作为step参数传递给random.randrange，以获得 15 的随机倍数。

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

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也可以使用pygame.Rects 进行碰撞检测并存储坐标。