Answer / munendra kumar

There are three acceptable answers: "Never," "Rarely," and "When the problem domain cannot be accurately modeled any other way."

There are some famous C++ pundits and luminaries who disagree with that third answer, but I will accept it.

Let's digress to consider this issue lest your interview turn into a religious debate. Consider an Asset class, Building class, Vehicle class, and CompanyCar class. All company cars are vehicles. Some company cars are assets because the organizations own them. Others might be leased. Not all assets are vehicles. Money accounts are assets. Real estate holdings are assets. Some real estate holdings are buildings. Not all buildings are assets. Ad infinitum. When you diagram these relationships, it becomes apparent that multiple inheritance is a likely and intuitive way to model this common problem domain. The applicant should understand, however, that multiple inheritance, like a chainsaw, is a useful tool that has its perils, needs respect, and is best avoided except when nothing else will do.

Answer / neelam saini

There are three acceptable answers:- "Never," "Rarely," and "When the problem domain cannot be accurately modeled any other way."

Consider an Asset class, Building class, Vehicle class, and CompanyCar class. All company cars are vehicles. Some company cars are assets because the organizations own them. Others might be leased. Not all assets are vehicles. Money accounts are assets. Real estate holdings are assets. Some real estate holdings are buildings. Not all buildings are assets. Ad infinitum. When you diagram these relationships, it becomes apparent that multiple inheritance is a likely and intuitive way to model this common problem domain. The applicant should understand, however, that multiple inheritance, like a chainsaw, is a useful tool that has its perils, needs respect, and is best avoided except when nothing else will do.

how can i access a direct (absolute, not the offset) memory address? here is what i tried: wrote a program that ask's for an address from the user, creates a FAR pointer to that adress and shows it. then the user can increment/decrement the value in that address by pressing p(inc+) and m(dec-). NOW, i compiled that program and opened it twice (in 2 different windows) and gave twice the same address to it. now look what happen - if i change the value in one "window" of the program, it DOES NOT change in the other! even if they point to the same address in the memory! here is the code snippet: //------------------------------------------------------ #include <stdio.h> //INCLUDE EVERY KNOWN HEADER FILE #include <conio.h> //FOR ANY CASE... #include <iostream.h> #include <dos.h> #include <process.h> main() { int far *ptr; //FAR POINTER!!! long address; char key=0; //A KEY FROM THE KEYBOARD int temp=0; clrscr(); cout<<"Enter Address:"; cin>>hex>>address; //GETS THE ADDRESS clrscr(); (long)ptr=address; temp=*ptr; //PUTS THE ADDRESS IN THE PTR cout<<"["<<hex<<(unsigned long)ptr<<"]="<<*ptr<<" = "<< (char)(*ptr); //SHOWS: [address]=value=ASCII symbol. while (key!=27) //WHILE YOU DONT PRESS ESC. { while(!kbhit()) //WHILE KEY IS NOT PRESSED { if (temp!=*ptr) { temp=*ptr; clrscr(); cout<<"["<<hex<< (unsigned long)ptr<<"]="<<*ptr<<" = "<<(char)(*ptr); }; //IF THE VALUE HAS CHANGED, CLEAR THE SCREEN AND SHOW //AGAIN if (key=='p') {key=0; (*ptr)++; } //INCREMENT VALUE if (key=='m') {key=0; (*ptr)--; } //DEC. VALUE }; key=getch(); //IF A KEY IS PRESSED, READ IT FROM THE //KEYBOARD }; return 0; //IF ESC WAS THE KEY, EXIT THE PROGRAM } //---------------------------------------------------------

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