我在C语言中使用SDL,我试图使一个字符在按下某个键时在屏幕上移动,并在释放该键时停止,但似乎在按住该键时触发了KEYUP事件。如果我在pressed()中删除KEYUP部分,字符将在屏幕上滑动,但显然不会自行停止。如果我将KEYUP部分保留在中,则必须反复按该键才能在屏幕上移动它们。
知道我做错了什么吗?
以下是我所拥有的:
...
int done = 0;
while (done==0)
{
pressed();
if (kenter == 1)
{
state = 1;
subscreen();
}
if (kescape == 1)
{
done = 1;
}
if (kup == 1)
{
playery += -2;
}
if (kdown == 1)
{
playery += 2;
}
if (kright == 1)
{
playerx += 2;
}
if (kleft == 1)
{
playerx += - 2;
}
...
int pressed ()
{
SDL_Event keyevent;
while (SDL_PollEvent(&keyevent))
{
switch(keyevent.type)
{
case SDL_KEYDOWN:
switch(keyevent.key.keysym.sym)
{
case SDLK_RETURN:
{
kenter = 1;
break;
}
case SDLK_ESCAPE:
{
kescape = 1;
break;
}
case SDLK_a:
{
ka = 1;
break;
}
case SDLK_s:
{
ks = 1;
break;
}
case SDLK_UP:
{
kup = 1;
break;
}
case SDLK_DOWN:
{
kdown = 1;
break;
}
case SDLK_RIGHT:
{
kright = 1;
break;
}
case SDLK_LEFT:
{
kleft = 1;
break;
}
default:
break;
}
}
switch(keyevent.type)
{
case SDL_KEYUP:
switch(keyevent.key.keysym.sym)
{
case SDLK_RETURN:
{
kenter = 0;
break;
}
case SDLK_ESCAPE:
{
kescape = 0;
break;
}
case SDLK_a:
{
ka = 0;
break;
}
case SDLK_s:
{
ks = 0;
break;
}
case SDLK_UP:
{
kup = 0;
break;
}
case SDLK_DOWN:
{
kdown = 0;
break;
}
case SDLK_RIGHT:
{
kright = 0;
break;
}
case SDLK_LEFT:
{
kleft = 0;
break;
}
default:
break;
}
}
}
return 0;
}
最佳答案
我想你的转换语句坏了。
用这种不那么混乱的方法
int pressed ()
{
SDL_Event event;
while(SDL_PollEvent(&event) )
{
if(event.type == SDLK_KEYDOWN)
{
switch(event.key.keysym.sym)
{
case SDLK_RETURN:
doStuff = 1
break;
default:
break;
}
}
if(event.type == SDLK_KEYUP)
{
switch(event.key.keysym.sym)
{
case SDLK_RETURN:
doStuff = 0;
break;
default:
break;
}
}
}
}
同样重要的是:
SDL Tutorials: Practical Keyboard Input
避免使用全局变量!
关于c - C SDL键盘事件SDL_KEYUP按下键时触发,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14102348/