对类成员的静态函数访问

Static function access to class member

本文关键字：静态函数 访问 成员      更新时间：2023-10-16

class Header:

#ifndef _APP_H_
#define _APP_H_
#include "glut.h"
#include "Declare.h"

class App {
private:
    static float angle; 
public:
    App();
    int OnExecute();
    void OnLoop();
    static void OnRender();
    bool OnInit();
    void OnCleanup();
    static void OnResize(int w, int h);

};

#endif
我对OnRender的定义是 
#include "App.h"
App::App() {
    angle = 0.0f;
}
void App::OnRender() {
    glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
    glLoadIdentity();
    gluLookAt( 0.0f, 0.0f, 10.0f,
               0.0f, 0.0f, 0.0f,
               0.0f, 1.0f, 0.0f);
    glRotatef(angle, 0.0f, 1.0f, 0.0f);
    glBegin(GL_TRIANGLES);
        glVertex3f(-2.0f,-2.0f,0.0f);
        glVertex3f(2.0f,0.0f,0.0f);
        glVertex3f(0.0f,2.0f,0.0f);
    glEnd();
    angle+=0.1f;
    glutSwapBuffers();
}
错误:

1>App.obj : error LNK2001: unresolved external symbol "private: static float App::angle" (?angle@App@@0MA)
1>App_OnRender.obj : error LNK2019: unresolved external symbol "private: static float App::angle" (?angle@App@@0MA) referenced in function "public: static void __cdecl App::OnRender(void)" (?OnRender@App@@SAXXZ)

与我如何在静态函数中引用静态变量有关。如果我不把角声明为static float angle那么我当然不能通过static void OnRender()来访问它。我必须补充更多的细节。如果我不将它声明为静态，就会得到这个错误illegal reference to non-static member App::angle

在源文件App.cpp中，您需要定义您的静态变量:

static float App::angle = 0; //0 is my test default value. use yours

，如果你想在非静态方法中使用angle，你可能想要使用App:angle的类实例。例如:

App::App() {
    App::angle = 0.0f;
}
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