Python实现贪吃蛇
时间: 2023-05-21 11:01:25 浏览: 107
可以使用 Pygame 模块来实现贪吃蛇游戏,以下是一个简单的 Python 代码示例:
```python
import pygame
import random
# 初始化 Pygame
pygame.init()
# 游戏窗口大小
window_width = 600
window_height = 400
# 定义颜色
white = (255, 255, 255)
black = (0, 0, 0)
red = (255, 0, 0)
# 创建游戏窗口
game_display = pygame.display.set_mode((window_width, window_height))
pygame.display.set_caption('贪吃蛇')
# 设置游戏时钟
clock = pygame.time.Clock()
# 定义蛇的初始位置和大小
snake_block_size = 10
snake_speed = 15
font_style = pygame.font.SysFont(None, 30)
def message(msg, color):
msg = font_style.render(msg, True, color)
game_display.blit(msg, [window_width/6, window_height/3])
def game_loop():
game_over = False
game_close = False
x1 = window_width / 2
y1 = window_height / 2
x1_change = 0
y1_change = 0
# 食物的初始位置
foodx = round(random.randrange(0, window_width - snake_block_size) / 10.0) * 10.0
foody = round(random.randrange(0, window_height - snake_block_size) / 10.0) * 10.0
while not game_over:
while game_close == True:
game_display.fill(white)
message("你输了!按 Q 退出,按 C 重新开始", red)
pygame.display.update()
for event in pygame.event.get():
if event.type == pygame.KEYDOWN:
if event.key == pygame.K_q:
game_over = True
game_close = False
if event.key == pygame.K_c:
game_loop()
for event in pygame.event.get():
if event.type == pygame.QUIT:
game_over = True
if event.type == pygame.KEYDOWN:
if event.key == pygame.K_LEFT:
x1_change = -snake_block_size
y1_change = 0
elif event.key == pygame.K_RIGHT:
x1_change = snake_block_size
y1_change = 0
elif event.key == pygame.K_UP:
y1_change = -snake_block_size
x1_change = 0
elif event.key == pygame.K_DOWN:
y1_change = snake_block_size
x1_change = 0
# 判断蛇是否撞到边界
if x1 >= window_width or x1 < 0 or y1 >= window_height or y1 < 0:
game_close = True
# 更新蛇的位置
x1 += x1_change
y1 += y1_change
# 绘制蛇和食物
game_display.fill(white)
pygame.draw.rect(game_display, black, [foodx, foody, snake_block_size, snake_block_size])
pygame.draw.rect(game_display, black, [x1, y1, snake_block_size, snake_block_size])
pygame.display.update()
# 判断蛇是否吃到食物
if x1 == foodx and y1 == foody:
print("Yummy!!")
clock.tick(snake_speed)
pygame.quit()
quit()
game_loop()
```
希望这个代码示例能够帮助你实现贪吃蛇游戏。