我怎么能随意在蛇身上挖洞呢？

Question

我有一条蛇，它在他身后旋转并画出轨迹，但我不知道如何在他画画的时候像随机停顿一样制造洞。我试过了，这不是随机的，但问题是它的停顿速度非常快。就像传送移动。这是我的代码：

import pygame
pygame.init()
import random
from random import randint

win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))

x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0

direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
    
    pygame.time.delay(5)
    drawing_time += 1


    if drawing_time == 1000:
        for pause in range(0, 60):
        
            head = pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 0)

            keys = pygame.key.get_pressed()

            if keys[pygame.K_LEFT]:
                direction.rotate_ip(-0.8)

            if keys[pygame.K_RIGHT]:
                direction.rotate_ip(0.8)

            x += direction.x
            y += direction.y
    
        time = 0

    keys = pygame.key.get_pressed()
    
    if keys[pygame.K_LEFT]:
        direction.rotate_ip(-0.8)

    if keys[pygame.K_RIGHT]:
        direction.rotate_ip(0.8)

    x += direction.x
    y += direction.y

    win.blit(background, (12020,0))
    head= pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
    pygame.display.update()

pygame.quit()

我试着给出头部半径0，也许这是个问题，但我不知道如何解决它。

Answer 1

不要在应用程序循环中实现控制游戏的循环。你需要应用程序循环。用它！

使用pygame.time.get_ticks()返回自调用pygame.init()以来的毫秒数。添加一个变量，指示是否需要绘制蛇(draw_snake)。定义一个变量，它指示需要更改状态的时间(next_change_time )。当时间已过期时，请更改变量的状态，并定义新的随机时间，在此时间必须再次更改状态。只有在设置变量时才画蛇。这会在蛇身上造成洞：

next_change_time = 0
draw_snake = False

run = True
while run:
    # [...]

    current_time = pygame.time.get_ticks()
    if current_time > next_change_time:
        next_change_time = current_time + random.randint(500, 1000)
        draw_snake = not draw_snake

    # [...]

    if draw_snake:
        pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)

    # [...]

使用pygame.time.Clock控制帧每秒，从而控制游戏的速度。

tick()对象的方法pygame.time.Clock以这种方式延迟游戏，循环的每一次迭代都会消耗相同的时间。

这意味着循环：

运行时

clock = pygame.time.Clock() run = True : clock.tick(60)

每秒跑60次。

完整的例子：

​

​

import pygame
pygame.init()
import random
from random import randint

win = pygame.display.set_mode((1280,720))
background = pygame.Surface(win.get_size(), pygame.SRCALPHA)
background.fill((0, 0, 0, 1))
clock = pygame.time.Clock()

x = randint(150,1130)
y = randint(150,570)
vel = 0.6
drawing_time = 0

next_change_time = 0
draw_snake = False

direction = pygame.math.Vector2(vel, 0).rotate(random.randint(0, 360))
run = True
while run:
    clock.tick(60)
    for event in pygame.event.get():
        if event.type == pygame.QUIT:
            run = False
    
    current_time = pygame.time.get_ticks()
    if current_time > next_change_time:
        next_change_time = current_time + random.randint(500, 1000)
        draw_snake = not draw_snake

    keys = pygame.key.get_pressed()
    if keys[pygame.K_LEFT]:
        direction.rotate_ip(-0.8)
    if keys[pygame.K_RIGHT]:
        direction.rotate_ip(0.8)

    x += direction.x
    y += direction.y

    win.blit(background, (12020,0))
    if draw_snake:
        pygame.draw.circle(win, (255,0,0), (round(x), round(y)), 5)
    pygame.display.update()

pygame.quit()

如果您想要为您需要的洞的其他大小，附加条件和不同的随机时间：

if current_time > next_change_time:
    draw_snake = not draw_snake
    if draw_snake:
        next_change_time = current_time + random.randint(500, 1000)
    else:
        next_change_time = current_time + random.randint(250, 500)