我正在编写一个 GLFW 应用程序,其中我将函数调用包装到一个简单的类中.我在设置键回调时遇到问题.我的班级定义为:
I'm writing a GLFW app, in which I've wrapped the function calls into a simple class. I'm having trouble setting the key callback. My class is defined as:
class GAME
{
private:
bool running;
public:
GAME();
int execute();
void events(int, int);
int loop();
int render();
};
执行函数是:
int GAME::execute()
{
glfwOpenWindow(640, 320, 8, 8, 8, 8, 0, 0, GLFW_WINDOW);
glfwSetWindowTitle("Viraj");
glfwSetKeyCallback(events);
running = true;
while(glfwGetWindowParam(GLFW_OPENED))
{
glfwPollEvents();
loop();
render();
}
return 0;
}
在 Visual Studio 2010 上编译以下代码会出现错误:
Compiling the following code on Visual Studio 2010 gives the error:
error C3867:'GAME::events':函数调用缺少参数列表;使用&GAME::events"创建一个指向成员的指针
使用 &GAME::events
给出:
error C2664: 'glfwSetKeyCallback' : 无法将参数 1 从 'void (__thiscall GAME::* )(int,int)' 转换为 'GLFWkeyfun' 1>没有上下文可以进行这种转换
有一个 C++ 语法用于指向类成员方法,但您不能将它们传递给 C 风格的 API.C 理解函数调用和每一个非静态对象方法,以你的 events
为例,在 C 语言中看起来是这样的:void events(void* this, int, int);
意味着除了标准参数之外的每个方法也会得到一个 this
指针,静默传递.
There is a C++ syntax for pointing to class member methods but you cannot pass them to a C style API. C understands function calls and every non-static object method, taking your events
as an example, looks like this thinking in C terms: void events(void* this, int, int);
meaning that every method apart from the standard arguments also gets a this
pointer silently passed.
为了让你的 events
C 兼容,让它static void events(int, int);
.这样它将遵循 C 调用语义 - 它不需要传递 this
指针.您还必须以某种其他方式将您的对象以某种方式传递给此回调(如果您在回调中需要此对象的数据).
To make your events
C compatible make it static void events(int, int);
. This way it will follow the C calling semantics - it will not require a this
pointer getting passed. You have to also somehow pass your object to this callback in some other manner (if you need this object's data in the callback).
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