最近一直在准备用来面试的几个小demo，为了能展现自己，所以都是亲自设计并实现的，其中一个就是在50行代码内来实现一个贪吃蛇，为了说明鄙人自己练习编程的一种方式--把代码写短，为了理解语言细节。 代码如下: <SPAN style="FONT-SIZE: 14px">import sys, pygame from pygame.locals import * from random import randrange up =lambda x:(x[0]-1,x[1]) down = lambda x :(x[0]+1,x[1]) left = lambda x : (x[0],x[1]-1) right = lambda x : (x[0],x[1]+1) tl = lambda x :x<3 and x+1 or 0 tr = lambda x :x==0 and 3 or x-1 dire = [up,left,down,right] move = lambda x,y:[y(x[0])]+x[:-1] grow = lambda x,y:[y(x[0])]+x s = [(5,5),(5,6),(5,7)] d = up food = randrange(0,30),randrange(0,40) FPSCLOCK=pygame.time.Clock() pygame.init() pygame.display.set_mode((800,600)) pygame.mouse.set_visible(0) screen = pygame.display.get_surface() screen.fill((0,0,0)) times=0.0 while True: time_passed = FPSCLOCK.tick(30) if times>=150: times =0.0 s = move(s,d) else: times +=time_passed for event in pygame.event.get(): if event.type == QUIT: sys.exit() if event.type == KEYDOWN and event.key == K_UP: s = move(s,d) if event.type == KEYDOWN and event.key == K_LEFT: d=dire[tl(dire.index(d))] if event.type == KEYDOWN and event.key == K_RIGHT: d=dire[tr(dire.index(d))] if s[0]==food: s = grow(s,d) food =randrange(0,30),randrange(0,40) if s[0] in s[1:] or s[0][0]<0 or s[0][0] >= 30 or s[0][1]<0 or s[0][1]>=40: break screen.fill((0,0,0)) for r,c in s: pygame.draw.rect(screen,(255,0,0),(c*20,r*20,20,20)) pygame.draw.rect(screen,(0,255,0),(food[1]*20,food[0]*20,20,20)) pygame.display.update()</SPAN>
