python - 基本运行周期不工作

Question

这个问题已经有答案了:

How do I pass a variable by reference? (41 个回答)

已关闭 7 年前。

我对 python 还很陌生，并且已经开始使用 pygame。我正在制作一款无尽的运行游戏，但我的运行周期遇到了问题。这是我的代码:

import pygame
import sys
from pygame.locals import *
pygame.init()
clock = pygame.time.Clock()
clock2 = pygame.time.get_ticks()
screen = pygame.display.set_mode((640, 575))
bgx = 0
bgx2 = -800
scroll = 10
pose = 1
background = pygame.image.load("images/background2.png").convert_alpha();
player = pygame.image.load("images/character1.png").convert_alpha();
screen.blit(background, (0, 0))
def draw_screen(x):
screen.blit(background, (x, 0))
def draw_screen2(x2):
screen.blit(background, (x2, 0))
def draw_player(pose):
    if pose == 1:
        player = pygame.image.load("images/character1.png").convert_alpha();
        screen.blit(player, (0, 0))
    elif pose == 2:
        player = pygame.image.load("images/character2.png").convert_alpha();
        screen.blit(player, (0, 0))
    elif pose == 3:
        player = pygame.image.load("images/character3.png").convert_alpha();
        screen.blit(player, (0, 0))
    elif pose == 4:
        player = pygame.image.load("images/character2.png").convert_alpha();
        screen.blit(player, (0, 0))
def set_poses(pose):
    if pose == 1:
        pose = 2
    if pose == 2:
        pose = 3
    if pose == 3:
        pose = 4
    if pose == 4:
        pose = 1

while True: #Loop
    clock2 = pygame.time.get_ticks()
    #Quitting Function
    for event in pygame.event.get():
        if event.type == QUIT:
            pygame.quit()
            sys.exit()
            quit()
    clock.tick(50)
    #Screen Scrolling
    if bgx != 800:
        bgx += scroll
    else:
        bgx = 0
    if bgx2 != 0:
        bgx2 += scroll
    else:
        bgx2 = -800
    #Drawing Items
    draw_screen2(bgx2)
    draw_screen(bgx)
    draw_player(pose)
    if clock2%6 == 0:
        set_poses(pose)
    pygame.display.update()

所以屏幕正在滚动，我的角色出现了，但他停留在它运行的第一个位置(我猜是框架。) set_poses(pose)我尝试让它在设置姿势之前打印一些东西，它确实做到了，但它没有设置姿势。我尝试让它打印姿势，结果表明姿势没有改变。所以，是的，我已经找到了我的问题，但我一生都找不到解决方案。

Answer 1

不要尝试通过引用传递 - 这就是您正在做的事情，而是尝试这样的事情:

def set_pose(pose):
    if pose >= 4:
        return 1
    else:
        return pose + 1

然后当您尝试在其他地方更新 pose 变量时:

pose = set_pose(pose)

您是否同意可以随意调用函数的参数？例如，set_pose 可能如下所示，并且执行完全相同相同的操作:

def set_pose(foo):
    if foo >= 4:
        return 1
    else:
        return foo + 1

换句话说，仅仅因为它在函数内部被称为 pose 并不意味着它会引用函数外部的pose。