Answer Posted / jknjkjk

Function overloading is like you have different functions
with the same name but different signatures working
differently. So, the compiler can differentiate and find
out which function to call depending on the context. In
case of operator overloading, you try to create your own
functions which are called when the corresponding operator
is invoked for the operands.

One important thing to understand is that you can create as
many functions as you want with the same name and sifferent
signatures so that they can work diffrently but for a
particular class, you cannot overload the operator function
based on number of arguments. There is a fundamental reason
behind this.

According to the rules, you can not create your own
operators but you have to use already available operators.
Another thing is since the operator are already defined for
use with buily-in types you can not change their
charecteristics. For example the binary operator '+' always
takes two parameters, so for this you cannot create a
function that takes three parameters. But you can always
overload them based on the type of the parameters.

What is the use of cmath in c++?

1073

---

How are pointers type-cast?

1117

---

What is a lambda function c++?

1113

---

What is class in c++ with example?

1277

---

Differentiate between the manipulator and setf( ) function?

1208

---

What is c++ manipulator?

989

---

Can you overload the operator+ for short integers?

1043

---

Explain deep copy and a shallow copy?

1067

---

Write a program to show polymorphism in C++?

1148

---

What is the purpose of the "delete" operator?

1065

---

int age=35; if(age>80) {Console.WriteLine("Boy you are old");} else {Console.WrieLine("That is a good age");}

1261

---

Is c# written in c++?

986

---

What is istream and ostream in c++?

1134

---

What is the exit function in c++?

989

---

What can I use instead of namespace std?

1156

---