python/pygame 在 windows 中的问题，但在 linux 中运行良好

如何解决python/pygame 在 windows 中的问题，但在 linux 中运行良好

好吧，所以我正在尝试使用 pygame 制作像 ludo 这样的游戏。当我在 Windows 上运行它时，播放器（蓝点）不会比第一行移动得更远，但是如果我在 linux 中运行它可以正常工作。 它根据玩家的位置绘制圆圈，但如果位置大于或等于 7，则不会打印。

这是游戏的代码（抱歉代码不好，谢谢！！）：

import random
import pygame

pygame.init()

black = ( 0,0)
white = ( 255,255,255)
blue = ( 0,255)

#cambia esta variable para cambiar la posicion de inicio
pos = 0

screenw = 600
screenh = 750
screen = pygame.display.set_mode((screenw,screenh))
cellsize = 75

tablero = ([0,0])

preg = (['tema: texto','S.O.','Edit texto','redes','edit imagen sonido y video','Edit hojas CÁLCULO','comodin para tema imagen','Infografía','Mapa conceptual','línea del tiempo','Captura de pantalla','Editar imagen','comodin para tema sonido','podcast PRESENTANDOSE','soundtrack (youtube .mp3)','Descargar de Ivoox','Editor sonido u editor online de extras','Descarga de soundsnap','comodin para tema video','VIDEO (youtube mp3)','Screen video & Desktop','Screen brower Tab','Webcam recording presentandose','Editar video u editor online','comodin para tema paginas web','Full screen captura','Editor webpage / editor online','cuaderno digital','HTML webpage','Firma en correo','has llegado al final!'])

dado = 0

tablero[pos] = 1


#draw the board
def tabla():
    pygame.draw.rect(screen,white,(cellsize,cellsize*6,cellsize,cellsize),3)
    pygame.draw.rect(screen,(cellsize*6,3)
    for x in range(1,8):
        for y in range(1,7):
            pygame.draw.line(screen,( x*cellsize,cellsize*6))
            pygame.draw.line(screen,( cellsize,y*cellsize),( cellsize*7,y*cellsize))
            pygame.display.update()

x = 0
clock = pygame.time.Clock()


#main loop
while x != 30 or pos >= 32:
    clock.tick(10)
    
    #draw the board
    tabla()

    #throw a dice 
    print("presione enter para lanzar dado")
    input()
    dado = random.randint(1,6)
    print("dado",dado)

    screen.fill(black)
    tabla()
 
    #remove the player and update the position
    tablero[pos] = 0
    preg[pos] = ' '
    pos += dado

    if pos >= 32:
        pos = 32
        tablero[pos] = 1
        print("enhorabuena,has ganado!! Puedes pedir tu punto extra de la evaluacion a Monica")
    #update the position
    tablero[pos] = 1
    print(tablero)
    print(pos)
    
    #this part isn't finished
    if pos == 4:
        for p in range(1,7):
            if preg[p] == ' ':
                screen.fill(black)
                tabla()
                print("elige una pregunta del bloque anterior antes de continuar")
                pos += 7

    #draw the circle that simbolices the player depending on the position
    #this is where I get the problem
    if pos >= pos and pos < 7:
        screen.fill(black)
        tabla()
        pygame.draw.circle(screen,blue,((cellsize*pos) + (cellsize/2),(cellsize*5 + (cellsize/2))),cellsize/3)

    elif pos >= 7 and pos < 13:
        screen.fill(black)
        tabla()
        pygame.draw.circle(screen,((cellsize*(pos-6)) + (cellsize/2),(cellsize*4 + (cellsize/2))),cellsize/3)

    elif pos >= 13 and pos < 19:
        screen.fill(black)
        tabla()
        pygame.draw.circle(screen,((cellsize*(pos-12)) + (cellsize/2),(cellsize*3 + (cellsize/2))),cellsize/3)

    elif pos >= 19 and pos < 25:
        screen.fill(black)
        tabla()
        pygame.draw.circle(screen,((cellsize*(pos-18)) + (cellsize/2),(cellsize*2 + (cellsize/2))),cellsize/3)

    elif pos >= 25 and pos < 32:
        screen.fill(black)
        tabla()
        pygame.draw.circle(screen,((cellsize*(pos-24)) + (cellsize/2),(cellsize + (cellsize/2))),cellsize/3)

    #print a question from a list depending on the player's position
    print("preg:",preg[pos - 1])

    #add 1 to the loop
    x += 1

解决方法

您必须处理应用程序循环中的事件。分别见pygame.event.get() pygame.event.pump()：

对于游戏的每一帧，您都需要对事件队列进行某种调用。这可确保您的程序可以在内部与操作系统的其余部分进行交互。

while x != 30 or pos >= 32:
    clock.tick(10)
    pygame.event.pump()

    # [...]