我用Pygame用python3做了一个经典的雅达利蛇游戏。我想派生一个子进程来监听击键,这样每当玩家输入一个键(向上、向下、向左或向右)时,子进程就会将该键发送给父进程。但是这条管道不应该被阻塞,这样蛇就可以沿着它行进的方向移动,直到收到密钥。
我在multi-processes上找到了Python的官方文档,但它没有描述我想要的行为,或者至少没有记录示例用法是否阻塞。有人能给我举个例子,说明如何实现这一点吗?
发布于 2019-01-16 16:58:41
你说过:
如果游戏的状态只是在每次迭代b/c时传递给AI,这是不公平的,那么它可能只需要计算下一步棋的时间。
KeyboardController处理键盘输入,而AsyncController启动一个线程并使用Queue类传递游戏状态和"AI“的决策。请注意,您必须在主线程上获取pygame事件,因此我在主循环中执行此操作,并简单地将事件向下传递给控制器。
您的AI必须由worker函数调用。正如您所看到的,目前worker函数中的"AI“仅每0.5秒执行一次操作,而帧率是120。人工智能需要这么长时间才能做出决定,这对游戏来说并不重要。
代码如下:
import pygame
import time
import random
from queue import Queue, Empty
from threading import Thread
class Controller():
def __init__(self, color, message, actor):
self.color = color
self.message = message
if actor: self.attach(actor)
def attach(self, actor):
self.actor = actor
self.actor.controller = self
self.actor.image.fill(self.color)
class AsyncController(Controller):
def __init__(self, actor=None):
super().__init__(pygame.Color('orange'), "AI is in control.", actor)
self.out_queue = Queue()
self.in_queue = Queue()
t = Thread(target=self.worker)
t.daemon = True
t.start()
def update(self, events, dt):
for e in events:
if e.type == pygame.KEYDOWN:
if e.key == pygame.K_SPACE: self.actor.controller = KeyboardController(self.actor)
self.out_queue.put_nowait((self.actor, events, dt))
try: return self.in_queue.get_nowait()
except Empty: pass
def worker(self):
while True:
try:
actor, events, dt = self.out_queue.get_nowait()
if actor.rect.x < 100: self.in_queue.put_nowait(pygame.Vector2(1, 0))
if actor.rect.x > 600: self.in_queue.put_nowait(pygame.Vector2(-1, 0))
if actor.rect.y < 100: self.in_queue.put_nowait(pygame.Vector2(0, 1))
if actor.rect.y > 400: self.in_queue.put_nowait(pygame.Vector2(0, -1))
if random.randrange(1, 100) < 15:
self.in_queue.put_nowait(random.choice([
pygame.Vector2(1, 0),
pygame.Vector2(-1, 0),
pygame.Vector2(0, -1),
pygame.Vector2(0, 1)]))
time.sleep(0.5)
except Empty:
pass
class KeyboardController(Controller):
def __init__(self, actor=None):
super().__init__(pygame.Color('dodgerblue'), "You're in control.", actor)
def update(self, events, dt):
for e in events:
if e.type == pygame.KEYDOWN:
if e.key == pygame.K_SPACE: self.actor.controller = AsyncController(self.actor)
if e.key == pygame.K_UP: return pygame.Vector2(0, -1)
if e.key == pygame.K_DOWN: return pygame.Vector2(0, 1)
if e.key == pygame.K_LEFT: return pygame.Vector2(-1, 0)
if e.key == pygame.K_RIGHT: return pygame.Vector2(1, 0)
class Actor(pygame.sprite.Sprite):
def __init__(self):
super().__init__()
self.image = pygame.Surface((32, 32))
self.image.fill(pygame.Color('dodgerblue'))
self.rect = self.image.get_rect(center=(100, 100))
self.direction = pygame.Vector2(1, 0)
self.pos = self.rect.center
def update(self, events, dt):
new_direction = self.controller.update(events, dt)
if new_direction:
self.direction = new_direction
self.pos += (self.direction * dt * 0.2)
self.rect.center = self.pos
def main():
pygame.init()
actor = Actor()
sprites = pygame.sprite.Group(actor)
screen = pygame.display.set_mode([800,600])
clock = pygame.time.Clock()
font = pygame.font.SysFont("consolas", 20, True)
dt = 0
KeyboardController(actor)
while True:
events = pygame.event.get()
for e in events:
if e.type == pygame.QUIT:
return
sprites.update(events, dt)
screen.fill(pygame.Color('grey12'))
screen.blit(font.render(actor.controller.message + ' [SPACE] to change to keyboard control.', True, pygame.Color('white')), (10, 10))
sprites.draw(screen)
dt = clock.tick(120)
pygame.display.update()
if __name__ == '__main__':
main()

请注意,此实现使用无限队列。您希望添加一些逻辑来清除队列,这样您的游戏就不会使用大量内存。
https://stackoverflow.com/questions/54209439
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