我使用fbo和vbo来绑定和绘制一个由Sprite.cpp中的两个三角形组成的矩形,下面是init方法
void Sprite::init(float x, float y, float width, float height) {
_x = x;
_y = y;
_width = width;
_height = height;
if (_vboID == 0) {
glGenBuffers(1, &_vboID);
}
vertexData[0] = x + width;
vertexData[1] = y + height;
vertexData[2] = x;
vertexData[3] = y + height;
vertexData[4] = x;
vertexData[5] = y;
vertexData[6] = x + width;
vertexData[7] = y + height;
vertexData[8] = x;
vertexData[9] = y;
vertexData[10] = x + width;
vertexData[11] = y;
glBindBuffer(GL_ARRAY_BUFFER, _vboID);
glBufferData(GL_ARRAY_BUFFER, sizeof(vertexData), vertexData, GL_DYNAMIC_DRAW);
glBindBuffer(GL_ARRAY_BUFFER, 0);}
这是抽签法
void Sprite::draw() {
glBindBuffer(GL_ARRAY_BUFFER, _vboID);
glEnableVertexAttribArray(0);
glVertexAttribPointer(0, 2, 0x1406, GL_FALSE, 0, 0);
glDrawArrays(GL_TRIANGLES, 0, 6);
glDisableVertexAttribArray(0);
glBindBuffer(GL_ARRAY_BUFFER, 0);}
我显然缺少一些非常明显的概念,因为我似乎找不到任何地方的解决方案,我的问题是,当我按下我的矩形不向下移动。(因为我使用的是WASD运动)下面是输入处理程序的代码
void InputHandler::handleInput(SDL_Event* event, Sprite *toMove) {
switch (event->type) {
case SDL_KEYDOWN:
{
if (event->key.keysym.sym == SDLK_a) {
toMove->setMovementBooleans(MovementTuples::LEFT, true);
std::cout << "Set left ";
}
if (event->key.keysym.sym == SDLK_d) {
toMove->setMovementBooleans(MovementTuples::RIGHT, true);
std::cout << "Set right ";
}
if (event->key.keysym.sym == SDLK_s) {
toMove->setMovementBooleans(MovementTuples::DOWN, true);
std::cout << "Set down ";
}
if (event->key.keysym.sym == SDLK_w) {
toMove->setMovementBooleans(MovementTuples::UP, true);
std::cout << "Set up ";
}
break;
}
case SDL_KEYUP:
{
if (event->key.keysym.sym == SDLK_a) {
toMove->setMovementBooleans(MovementTuples::LEFT, false);
}
if (event->key.keysym.sym == SDLK_d) {
toMove->setMovementBooleans(MovementTuples::RIGHT, false);
}
if (event->key.keysym.sym == SDLK_s) {
toMove->setMovementBooleans(MovementTuples::DOWN, false);
}
if (event->key.keysym.sym == SDLK_w) {
toMove->setMovementBooleans(MovementTuples::UP, false);
}
}
}}
静态访问此方法。以下是在我的sprite cpp中单击s键时移动sprite的代码
void Sprite::update() {
if (movementBooleans[MovementTuples::DOWN]) {
_x -= .1f;
init(_x, _y, _width, _height);
std::cout << "Moved to: (" << _x << ", " << _y << ")"<< std::endl;
}当我点击向下,我的矩形完全消失,_x值保持不变。我测试了输入,它的工作方式与我想要的完全一样,我只是不能让矩形移动。有人能帮我弄清楚怎么移动雪碧吗,谢谢!
发布于 2016-08-08 22:52:28
我在用OpenGL。虽然我没有使用SDL,因为我使用的是用户实现的输入处理程序,但概念应该是相同的。
我目前使用的Game有一个Game对象,它是从Engine类对象继承的,其中Engine类是Singleton对象。由于这个游戏引擎的结构,引擎类需要游戏类来实现一个虚拟的keyboardInput函数。这是原型的样子
引擎类
class Engine : public Singleton {
protected:
// Protected Members
private:
// Private Members
public:
// virtual destructor & public functions
protected:
// explicit protected Constructor & protected functions
private:
bool messageHandler( unsigned uMsg, WPARAM wParam, LPARAM lParam );
virtual void keyboardInput( unsigned vkCode, bool isPressed ) = 0;
// Other Private Functions
}; // Engine游戏类
class Game sealed : public Engine {
private:
// private members here
public:
// Constructor and virtual Destructor
private:
virtual void keyBoardInput( unsigned vkCode, bool isPressed ) override;
}; // Game下面是用于Windows的messageHandler()函数
// ----------------------------------------------------------------------------
// messageHandler()
bool Engine::messageHandler( unsigned uMsg, WPARAM wParam, LPARAM lParam ) {
switch( uMsg ) {
case WM_CLOSE: {
PostQuitMessage( 0 );
return true;
}
case WM_SYSKEYDOWN : {
if ( ( VK_MENU == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_RMENU; // Alt Key
}
// Fall Through
}
case WM_KEYDOWN: {
if ( ( VK_RETURN == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_SEPARATOR;
} else if ( ( VK_CONTROL == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_RCONTROL;
}
if ( 0 == ( lParam & 0x40000000 ) ) { // Supress Key Repeats
keyboardInput( wParam, true );
}
return true;
}
case WM_SYSKEYUP: {
if ( ( VK_MENU == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_RMENU; // Alt Key
}
// Fall Through
}
case WM_KEYUP: {
if ( ( VK_RETURN == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_SEPARATOR;
} else if ( ( VK_CONTROL == wParam ) && ( lParam & 0x1000000 ) ) {
wParam = VK_RCONTROL;
}
keyboardInput( wParam, false );
return true;
}
case WM_MOUSEMOVE: {
// Mouse Motion Detected, Coordinates Are WRT Window Therefore
// 0,0 Is The Coordinate Of The Top Left Corner Of The Window
m_mouseState.position = glm::ivec2( LOWORD( lParam ), HIWORD( lParam ) );
mouseInput();
return true;
}
case WM_LBUTTONDOWN: {
m_mouseState.isButtonPressed[MOUSE_LEFT_BUTTON] = true;
mouseInput();
return true;
}
case WM_LBUTTONUP: {
m_mouseState.isButtonPressed[MOUSE_LEFT_BUTTON] = false;
mouseInput();
return true;
}
case WM_RBUTTONDOWN: {
m_mouseState.isButtonPressed[MOUSE_RIGHT_BUTTON] = true;
mouseInput();
return true;
}
case WM_RBUTTONUP: {
m_mouseState.isButtonPressed[MOUSE_RIGHT_BUTTON] = false;
mouseInput();
return true;
}
case WM_MBUTTONDOWN: {
m_mouseState.isButtonPressed[MOUSE_MIDDLE_BUTTON] = true;
mouseInput();
return true;
}
case WM_MBUTTONUP: {
m_mouseState.isButtonPressed[MOUSE_MIDDLE_BUTTON] = false;
mouseInput();
return true;
}
case WM_MOUSEWHEEL: {
// Mouse Wheel Moved
// wParam Contains How Much It Was Moved
return true;
}
default: {
return false; // Did Not Handle The Message
}
}
} // messageHandler最后,这是keyboardInput()函数
// ----------------------------------------------------------------------------
// keyboardInput()
void Game::keyboardInput( unsigned vkCode, bool isPressed ) {
std::ostringstream strStream;
strStream << "Key 0" << std::hex << vkCode << " was " << ( isPressed ? "Pressed" : "Released" );
Logger::log( strStream );
if ( VK_ESCAPE == vkCode ) {
PostQuitMessage( 0 );
}
static bool keyPressed[256] = { 0 };
if ( vkCode < 256 ) {
keyPressed[vkCode] = isPressed;
}
if ( isPressed ) {
return;
}
switch ( vkCode ) {
case VK_DOWN:
case 'S' : {
// do logic here
break;
}
case VK_LEFT:
case 'A' : {
// do logic here
break;
}
case VK_RIGHT:
case 'D' : {
// do logic here
break;
}
case VK_UP:
case 'W' : {
// do logic here
break;
}
}
} // handleKeyboard当您仔细查看代码时,我不是在轮询键何时被按下,而是在查询或等待密钥何时释放。这个逻辑的原因是这样的:当你按下一个键时,它可以被反复按下,而不会触及向上状态。这就像在任何文本编辑器中按下相同的键,就会看到屏幕上显示相同的键。避免这种行为;我们做的是相反的逻辑。
我们寻找钥匙何时被释放。密钥只能释放一次,必须再次按下才能再次进入释放状态。这样你就可以让程序的动作发生一次,每次按下键,然后释放。
现在,如果您注意到,我们检查键是否处于按下状态,然后返回;这很重要,因为如果不这样做,程序将自动执行消息处理程序中的操作,因为默认情况下键按状态已经处于释放状态。因此,在调用Game::keyboardInput()和Engine::messageHandler()函数时,在呈现框架或更新框架期间发生了什么,如果它检测到按键已将其状态更改为true,则从该函数返回并再次调用,直到释放密钥;一旦释放密钥,状态更改为已释放的状态,然后跳过if语句,直接转到正在处理的键的开关语句。
发布于 2016-08-09 16:05:14
如果你想移动一个OpenGL形状,你真的应该做矩阵转换。
我个人使用GLM为我做所有的繁重工作,所以我所要做的就是在转换完成后将矩阵传递给GLSL。
https://stackoverflow.com/questions/38839279
复制相似问题