GLFW lambdas可以接受捕获参数吗？

Question

glfwSetKeyCallback(window, 
        [this](GLFWwindow* window, int key, int scancode, int action, int mods)
        {
            if(action == GLFW_PRESS)
            {
                 //use a mutex
                 //Modify member variable
            }
        });

不存在从“lambda[]void（GLFWwindow*window,int key,int scancode,int action,int mods）->void”到glfwkeyfun“的适当转换函数

我是漏掉了什么还是这段代码只是无效？

Answer 1

GLFW回调不接受lambdas或函数对象：它们接受普通的旧函数指针。非捕获的λ可以转换为函数指针，但不能转换为捕获的。

但是，通过使用glfwsetuserpointer和glfwgetuserpointer可以获得类似的效果。lambda仍然不能捕获，但是您可以恢复this指针。

例如，

struct MyClass {
  GLFWwindow* window;

  MyClass(GLFWwindow* window) : window(window) {
    glfwSetWindowUserPointer(window, static_cast<void*>(this));

    glfwSetKeyCallback(window, 
      [](GLFWwindow* window, int key, int scancode, int action, int mods) {
        auto self = static_cast<MyClass*>(glfwGetWindowUserPointer(window));
        // can access member variables through `self`
      });
  }

  // make sure that either the class will last as long as GLFW will
  // or clean up the user pointer and callbacks in here
  ~MyClass() {
    // clean up
  }

  // don't be able to copy, probably, or bad things will happen
  MyClass(const MyClass&) = delete;
  MyClass& operator=(const MyClass&) = delete;
  // other things...
};