如何解决Pong - 当任一玩家达到分数阈值时崩溃而不是重置游戏
问题是每次显示获胜者文本后游戏都会崩溃（如果任何一个玩家得分达到 10 或更高），并且当我声明它应该时无法重新调用 main() 函数，正如你所看到的我的代码如下。因此，当它显示获胜者文本时，它应该暂停游戏 5 秒，然后退出 while 循环并调用 main() 但它在获胜者文本显示后立即崩溃（参见下面的屏幕截图）。获胜者文本的逻辑在我的 main() 函数的顶部附近。
不确定根本原因是什么，因为我在程序顶部调用了 pygame.init()。任何帮助将不胜感激！完整的代码如下......如果有人可以运行它并让我知道它崩溃的根本原因是什么，那就太好了。提前致谢！
import pygame
pygame.init()
WIDTH,HEIGHT = 750,500
WINDOW = pygame.display.set_mode((WIDTH,HEIGHT))
pygame.display.set_caption("PONG")
WINNER_FONT = pygame.font.SysFont('comicsans',100)
BLUE = (0,255)
RED = (255,0)
YELLOW = (255,255,0)
WHITE = (255,255)
BLACK = (0,0)
class Paddle(pygame.sprite.Sprite):
def __init__(self):
pygame.sprite.Sprite.__init__(self)
self.image = pygame.Surface([10,75])
self.image.fill(WHITE)
self.rect = self.image.get_rect()
self.paddle_speed = 4
self.points = 0
class Ball(pygame.sprite.Sprite):
def __init__(self):
pygame.sprite.Sprite.__init__(self)
self.image = pygame.Surface([10,10])
self.image.fill(YELLOW)
self.rect = self.image.get_rect()
self.speed = 2
self.dx = 1
self.dy = 1
paddle1 = Paddle()
paddle1.image.fill(BLUE)
paddle1.rect.x = 25
paddle1.rect.y = 225
paddle2 = Paddle()
paddle2.image.fill(RED)
paddle2.rect.x = 715
paddle2.rect.y = 225
ball = Ball()
ball.rect.x = 375
ball.rect.y = 250
all_sprites = pygame.sprite.Group()
all_sprites.add(paddle1,paddle2,ball)
def redraw():
WINDOW.fill(BLACK)
font = pygame.font.SysFont("Consolas",35)
text = font.render("---[PONG]---",1,YELLOW)
textRect = text.get_rect()
textRect.center = (WIDTH//2,25)
WINDOW.blit(text,textRect)
#Player 1 score
p1_score = font.render(str([paddle1.points]),WHITE)
p1Rect = p1_score.get_rect()
p1Rect.center = (50,50)
WINDOW.blit(p1_score,p1Rect)
#Player 2 score
p2_score = font.render(str([paddle2.points]),WHITE)
p2Rect = p2_score.get_rect()
p2Rect.center = (700,50)
WINDOW.blit(p2_score,p2Rect)
all_sprites.draw(WINDOW)
pygame.display.update()
def draw_winner(text):
draw_text = WINNER_FONT.render(text,WHITE)
WINDOW.blit(draw_text,(WIDTH//2 - draw_text.get_width()/2,HEIGHT//2 - draw_text.get_height()/2))
pygame.display.update()
pygame.time.delay(5000)
def main():
run = True
while run:
pygame.time.delay(10)
for event in pygame.event.get():
if event.type == pygame.QUIT:
run = False
pygame.quit()
winner_text = ""
if paddle1.points >= 10:
winner_text = "Blue Wins!"
if paddle2.points >= 10:
winner_text = "Red Wins!"
if winner_text != "":
draw_winner(winner_text)
break
# Paddle movement
key = pygame.key.get_pressed()
if key[pygame.K_w]:
paddle1.rect.y -= paddle1.paddle_speed
if key[pygame.K_s]:
paddle1.rect.y += paddle1.paddle_speed
if key[pygame.K_UP]:
paddle2.rect.y -= paddle2.paddle_speed
if key[pygame.K_DOWN]:
paddle2.rect.y += paddle2.paddle_speed
# Ensures paddles never move off the screen
if paddle1.rect.y < 0:
paddle1.rect.y = 0
if paddle1.rect.y > 425:
paddle1.rect.y = 425
if paddle2.rect.y < 0:
paddle2.rect.y = 0
if paddle2.rect.y > 425:
paddle2.rect.y = 425
# Ball movement
ball.rect.x += ball.speed * ball.dx
ball.rect.y += ball.speed * ball.dy
# Setting up collision detection with the walls by changing ball's direction
if ball.rect.y > 490:
ball.dy = -1
if ball.rect.x > 740:
ball.rect.x,ball.rect.y = 375,250
paddle1.points += 1
if ball.rect.y < 10:
ball.dy = 1
if ball.rect.x < 10:
ball.rect.x,250
paddle2.points += 1
# Setting up collision detection with the paddles
if paddle1.rect.colliderect(ball.rect):
ball.dx = 1
if paddle2.rect.colliderect(ball.rect):
ball.dx = -1
redraw()
main()
if __name__ == "__main__":
main()
解决方法
永远不要递归调用应用程序循环，也不要在应用程序循环中“等待”。请参阅 How to wait some time in pygame? 和 Python : What is the better way to make multiple loops in pygame?。
创建一个 init 函数来初始化所有游戏状态：
def init():
global paddle1,paddle2,ball,all_sprites
paddle1 = Paddle()
paddle1.image.fill(BLUE)
paddle1.rect.x = 25
paddle1.rect.y = 225
paddle2 = Paddle()
paddle2.image.fill(RED)
paddle2.rect.x = 715
paddle2.rect.y = 225
ball = Ball()
ball.rect.x = 375
ball.rect.y = 250
all_sprites = pygame.sprite.Group()
all_sprites.add(paddle1,ball)
需要重新启动游戏时调用init函数：
def main():
run = True
while run:
# [...]
if winner_text != "":
draw_winner(winner_text)
init()
# [...]
# main() <--- DELETE
添加 game_over 状态并使用 pygame.time.get_ticks() 获取自调用 pygame.init() 以来的毫秒数。游戏结束时，设置状态game_over并计算游戏必须重新开始的时间。时机成熟时，重置 game_over 并调用 init。根据{{1}}绘制场景：
game_over使用 pygame.time.Clock 控制每秒帧数，从而控制游戏速度。
tick() 对象的方法 pygame.time.Clock 以这种方式延迟游戏，即循环的每次迭代消耗相同的时间段。见pygame.time.Clock.tick()：
这个方法应该每帧调用一次。
这意味着循环：
def main():
init()
winner_text = ""
restart_time = 0
game_over = False
clock = pygame.time.Clock()
run = True
while run:
clock.tick(100)
current_time = pygame.time.get_ticks()
# [...]
if not game_over:
if paddle1.points >= 10:
winner_text = "Blue Wins!"
restart_time = current_time + 5000
game_over = True
if paddle2.points >= 10:
winner_text = "Red Wins!"
restart_time = current_time + 5000
game_over = True
if game_over:
draw_winner(winner_text)
if current_time > restart_time:
init()
game_over = False
else:
continue
# [...]
每秒运行 100 次。
完整示例：
clock = pygame.time.Clock()
run = True
while run:
clock.tick(100)
首先让我们深入了解一下您的 main() 函数：
def main():
run = True
while run:
#[...]some part of code here
if winner_text != "":
draw_winner(winner_text)
break
#[...]some part of code here
main()
现在只要有人赢了 paddle1.points = 10 或 paddle2.points = 10
if paddle1.points >= 10:
winner_text = "Blue Wins!"
if paddle2.points >= 10:
winner_text = "Red Wins!"
if winner_text != "":
draw_winner(winner_text)
break
现在只要有人赢了，你的这部分代码就会被执行：
if winner_text != "":
draw_winner(winner_text)
break
因此，您只是显示“蓝色获胜”或“红色获胜”之类的文本，然后您会跳出 while 循环并再次调用 main() 函数，但您并未重置 { {1}} 和 paddle1.points 分别为 0。
所以您的 paddle2.points 函数应该类似于：
main()这是完整的工作代码：
def main():
run = True
paddle1.points = 0
paddle2.points = 0
while run:
#[...]while loop here
虽然这可行，但我建议您采用@Rabbid76 所说的，因为递归调用应用程序循环从来不被认为是一种好的做法
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