In Python, how to make the user wait before pressing another key?

Question

I'm coding a simple snake game in pygame. The thing is, if the user presses UP and LEFT (just an example) too fast, the snake will collide with itself (it will not have the time to go up before trying to go left, thus colliding with its own body and dying).

I figured it could be solved if the user had to wait a quarter of a second or so before being able to move the snake again, so the snake would always have time to go up one step before changing directions. But how do I do it?

I've already tried simply reduncing the framerate of my game, but the problem remains.

    import pygame
from pygame.locals import *
from sys import exit
from random import randint

pygame.init()

largura = 640
altura = 480
x_cobra = int(largura/2 - 20)
y_cobra = int(altura/2 - 25)
x_maca = randint(40,600)
y_maca = randint(50, 430)
pontos = 0
fonte = pygame.font.SysFont('arial', 40, True, True)
tela = pygame.display.set_mode((largura, altura))
pygame.display.set_caption('Jogo da cobrinha')
relogio = pygame.time.Clock()
lista_cobra = []
comprimento_inicial = 5
velocidade = 20
x_controle = velocidade
y_controle = 0
morreu = False

def aumenta_cobra(lista_cobra):
    for XeY in lista_cobra:
        pygame.draw.rect(tela, (0,255,0), (XeY[0],XeY[1], 20,20))

def reiniciar_jogo():
    global pontos, comprimento_inicial, x_cobra, y_cobra, lista_cobra, lista_cabeca, x_maca, y_maca, morreu
    pontos = 0
    comprimento_inicial = 5
    x_cobra = int(altuara/2)
    y_cobra = int(largura/2)
    lista_cobra = []
    lista_cabeca = []
    x_maca = randint(40,600)
    y_maca = randint(50,430)
    morreu = False

while True:
    relogio.tick(15)
    tela.fill((255,255,255))
    mensagem = f'Pontos: {pontos}'
    texto_formatado = fonte.render(mensagem, True, (0,0,0))
    for event in pygame.event.get():
        if event.type == QUIT:
            pygame.quit()
            exit()

        if event.type == KEYDOWN:
            if event.key == K_LEFT and x_controle != velocidade:
                x_controle = -velocidade
                y_controle = 0
            if event.key == K_DOWN and y_controle != -velocidade:
                x_controle = 0
                y_controle = velocidade
            if event.key == K_UP and y_controle != velocidade:
                x_controle = 0
                y_controle = -velocidade
            if event.key == K_RIGHT and x_controle != -velocidade:
                x_controle = velocidade
                y_controle = 0

    x_cobra = x_cobra + x_controle
    y_cobra = y_cobra + y_controle

    cobra = pygame.draw.rect(tela, (0,255,0), (x_cobra, y_cobra,20,20))
    maca = pygame.draw.rect(tela, (255,0,0), (x_maca, y_maca,20,20))

    if cobra.colliderect(maca):
        x_maca = randint(40, 600)
        y_maca = randint(50, 430)
        pontos += 1
        comprimento_inicial += 1

    lista_cabeca = []
    lista_cabeca.append(x_cobra)
    lista_cabeca.append(y_cobra)
    lista_cobra.append(lista_cabeca)

    if lista_cobra.count(lista_cabeca) > 1:
        morreu = True
        while morreu:
            for event in pygame.event.get():
                if event.type == QUIT:
                    pygame.quit()
                    exit()
                if event.type == KEYDOWN:
                    if event.key == K_r:
                        reiniciar_jogo()

    if len(lista_cobra) > comprimento_inicial:
        del lista_cobra[0]

    aumenta_cobra(lista_cobra)


    tela.blit(texto_formatado, (400, 30))
    pygame.display.update()

Can you include the minimum code required to re-create the issue? — Mehrdad Moradi, Sep 07 '22 at 17:12
Safe the time and don't handle any keyboard events until time + x. — Klaus D., Sep 07 '22 at 17:13
Just don't allow a move that would cause the head of the snake to collide with the second segment. No timeout required. — kindall, Sep 07 '22 at 17:23

score 1 · Answer 1 · answered Sep 07 '22 at 17:20

Just ignore any new key press events until you have updated the model for the previous key press. You could use some kind of flag in your key press event handler that is set by the key press and reset by the logic that draws the snake. If the flag is true then do nothing.

score 0 · Accepted Answer · answered Sep 07 '22 at 17:35

One way to solve this is to just only accept the first input on each movement frame. So instead of allowing the event loop to run for all inputs, it stops and returns the first valid input.

Something like this (see get_keys function):

import pygame
from pygame.locals import *
from sys import exit
from random import randint

pygame.init()

largura = 640
altura = 480
x_cobra = int(largura/2 - 20)
y_cobra = int(altura/2 - 25)
x_maca = randint(40,600)
y_maca = randint(50, 430)
pontos = 0
fonte = pygame.font.SysFont('arial', 40, True, True)
tela = pygame.display.set_mode((largura, altura))
pygame.display.set_caption('Jogo da cobrinha')
relogio = pygame.time.Clock()
lista_cobra = []
comprimento_inicial = 5
velocidade = 20
x_controle = velocidade
y_controle = 0
morreu = False

def aumenta_cobra(lista_cobra):
    for XeY in lista_cobra:
        pygame.draw.rect(tela, (0,255,0), (XeY[0],XeY[1], 20,20))

def reiniciar_jogo():
    global pontos, comprimento_inicial, x_cobra, y_cobra, lista_cobra, lista_cabeca, x_maca, y_maca, morreu
    pontos = 0
    comprimento_inicial = 5
    x_cobra = int(altuara/2)
    y_cobra = int(largura/2)
    lista_cobra = []
    lista_cabeca = []
    x_maca = randint(40,600)
    y_maca = randint(50,430)
    morreu = False

def get_keys():
    for event in pygame.event.get():
        if event.type == QUIT:
            pygame.quit()
            exit()

        if event.type == KEYDOWN:
            if event.key == K_LEFT and x_controle != velocidade:
                # x_controle = -velocidade
                # y_controle = 0
                return [-velocidade,0]
            if event.key == K_DOWN and y_controle != -velocidade:
                # x_controle = 0
                # y_controle = velocidade
                return [0, velocidade]
            if event.key == K_UP and y_controle != velocidade:
                # x_controle = 0
                # y_controle = -velocidade
                return [0, -velocidade]
            if event.key == K_RIGHT and x_controle != -velocidade:
                # x_controle = velocidade
                # y_controle = 0
                return [velocidade, 0]
    return [x_controle, y_controle] # If no valid key presses found it continues moving in the same direction.

while True:
    relogio.tick(15)
    tela.fill((255,255,255))
    mensagem = f'Pontos: {pontos}'
    texto_formatado = fonte.render(mensagem, True, (0,0,0))
    
    [x_controle, y_controle] = get_keys()

    x_cobra = x_cobra + x_controle
    y_cobra = y_cobra + y_controle

    cobra = pygame.draw.rect(tela, (0,255,0), (x_cobra, y_cobra,20,20))
    maca = pygame.draw.rect(tela, (255,0,0), (x_maca, y_maca,20,20))

    if cobra.colliderect(maca):
        x_maca = randint(40, 600)
        y_maca = randint(50, 430)
        pontos += 1
        comprimento_inicial += 1

    lista_cabeca = []
    lista_cabeca.append(x_cobra)
    lista_cabeca.append(y_cobra)
    lista_cobra.append(lista_cabeca)

    if lista_cobra.count(lista_cabeca) > 1:
        morreu = True
        while morreu:
            for event in pygame.event.get():
                if event.type == QUIT:
                    pygame.quit()
                    exit()
                if event.type == KEYDOWN:
                    if event.key == K_r:
                        reiniciar_jogo()

    if len(lista_cobra) > comprimento_inicial:
        del lista_cobra[0]

    aumenta_cobra(lista_cobra)


    tela.blit(texto_formatado, (400, 30))
    pygame.display.update()

Worked! Thank you! But I don't really understand why. Is it because running the rest of the code before checking for other inputs takes a while? — Matheus S., Sep 07 '22 at 23:32
The `get_keys()` function returns after the first valid key down event, so additional key down events are ignored. — import random, Sep 08 '22 at 12:52
@MatheusS. it's because the events list can have more than one keypress stored in it, so returning on the first one means the extras are ignored. It does have the side effect that if you press an invalid key and a valid key in an unfortunate order it may disregard the movement entirely. I'd recommend using kindall's suggestion of simply not allowing a move that would impact the second segment. But if you're satisfied with the solution then please mark it as answered, thanks. — Jackson V, Sep 09 '22 at 00:31

In Python, how to make the user wait before pressing another key?

2 Answers2