Question

我有这个：

// Get the Application Constructor Vector
vector<ApplicationConstructor> Application::getAppConstructorVector(){
   return this->AppConstructorVector;
}

此处Application是一个抽象类，带有一个名为ShowData()的抽象方法：

 void DesktopApplication::ShowData() const{
 cout << "-------------------------AppConstructor Details----------------------------------------------" << endl;
 vector<ApplicationConstructor> appConstructorVector = this->getAppConstructorVector();
 for (int i= 0; i < appConstructorVector.size(); i ++)
 {
     cout << "Contructor Code: " << appConstructorVector.at(i).getConstructorCode()
          << "Name: " << appConstructorVector.at(i).getConstructorName() 
          << "Email: " << appConstructorVector.at(i).getConstructorEmail()
          << endl;
 }     
}

其中desktopApplication扩展了Application并实现了抽象方法Showdata()

我有这个：

void AppSystem::ShowData(const Application &app){
   app.ShowData();
}

其中AppSystem是具有vector<Application *> &ApplicationVector;属性的类基本上我有这个

  GameApplication app1(true, "Puzzle", (char *)"game1456", "Tetris", "1.0.0", appConstructor1, 10.0); 
  GameApplication *appPtr1 = &app1;
  AppSystem appSystem;
  appSystem.PushAppToApplicationVector(appPtr1);
  appSystem.ShowData(*appPtr1);

我明白了：

错误：将“ const DesktopApplication”作为“ this”参数传递限定词[-fpermissive]尽管我称GameApplication showData（）不是DesktopApplication之一。有什么建议吗？

将'const'传递为'this'参数将丢弃限定符[-fpermissive]

0 个答案: