Pygame - “错误:显示 Surface 退出" [英] Pygame - "error: display Surface quit"

问题描述

我制作了一个菜单屏幕，点击按钮会在同一窗口中打开不同的屏幕.

def main():导入 pygame，随机，时间pygame.init()大小=[800, 600]屏幕=pygame.display.set_mode(大小)pygame.display.set_caption("游戏")完成=假时钟=pygame.time.Clock()完成时==假:对于 pygame.event.get() 中的事件:pos = pygame.mouse.get_pos()如果 event.type == pygame.QUIT:完成=真休息如果 button_VIEW.collidepoint(pos):如果 event.type == pygame.MOUSEBUTTONDOWN:打印(查看.")看法()休息屏幕填充(黑色)...定义视图():完成=假时钟=pygame.time.Clock()完成时==假:对于 pygame.event.get() 中的事件:pos = pygame.mouse.get_pos()如果 event.type == pygame.QUIT:完成=真休息...

如果可能，我想知道如何避免错误:

 screen.fill(black)错误:显示 Surface 退出>>>

查看此处的其他问题后，我尝试将 breaks 添加到任何循环的出口，但仍然发生错误.

我理解问题是程序试图在窗口关闭后执行 screen.fill(black)，但我没有关于如何防止错误的进一步想法.>

感谢您的帮助.对不起，如果它看起来很简单.

解决方案

几种可能性:

• 在 view 函数中结束进程(例如使用 sys.exit()).不理想.
• 从 view 函数返回一个值来指示应用程序应该结束(例如 return done)，并在 main<中检查该返回值/code> 函数(如果完成:返回).更好.
• 使 done 成为全局的，并在 main 函数中检查它的值.我真的不喜欢这个解决方案.
• 我的最爱:完全避免多个事件循环，这样问题就会自行解决(这样你就可以例如退出 main 函数并返回).

I've made a menu screen where clicking on a button leads to a different screen in the same window.

def main():
    import pygame, random, time
    pygame.init()

    size=[800, 600]
    screen=pygame.display.set_mode(size)
    pygame.display.set_caption("Game")
    done=False
    clock=pygame.time.Clock()

    while done==False:
        for event in pygame.event.get():
            pos = pygame.mouse.get_pos()
            if event.type == pygame.QUIT:
                done=True
                break
            if button_VIEW.collidepoint(pos):
                if event.type == pygame.MOUSEBUTTONDOWN:
                    print("VIEW.")
                    view()
                    break

         screen.fill(black)
            ...

def view():
    done=False
    clock=pygame.time.Clock()

    while done==False:
        for event in pygame.event.get():
            pos = pygame.mouse.get_pos()
            if event.type == pygame.QUIT:
                done=True
                break
            ...

If possible, I'd like to know how I can avoid the error:

    screen.fill(black)
error: display Surface quit
>>> 

After looking at other questions on here, I tried adding breaks to the exits of any loops, but still the error occurs.

I understand the issue is that the program is trying to execute screen.fill(black) after the window has been closed, but I have no further ideas on how to prevent the error.

I appreciate any help. Sorry if it seems simple.

解决方案

Several possibilities:

• end the process (with e.g. sys.exit()) in the view function. Not ideal.
• return a value from the view function to indicate that the application shoud end (e.g. return done), and check for that return value in the main function (if done: return). Better.
• make done global and check for its value in the main function. I really would not like this solution.
• my favourite: avoid multiple event loops altogether, so the problem solves itself (so you could just e.g. exit the main function with return).

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