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2. The Three Basic Rules of Operator Overloading in C++ - Stack...

   stackoverflow.com/.../what-are-the-basic-rules-and-idioms-for-operator-overloading

   The function call operator, used to create function objects, also known as functors, must be defined as a member function, so it always has the implicit this argument of member functions. Other than this, it can be overloaded to take any number of additional arguments, including zero.

3. overriding - Override and overload in C++ - Stack Overflow

   stackoverflow.com/questions/429125

   std::wstring ToUnicode(std::string s); In some cases it's worth arguing that a function of a different name is a better choice than an overloaded function. In the case of constructors, overloading is the only choice. Over riding a function is entirely different, and serves an entirely different purpose.

4. What is the use/advantage of function overloading?

   stackoverflow.com/questions/3343913

   7. Overloading is a form of polymorphism. It allows the programmer to write functions to do conceptually the same thing on different types of data without changing the name. (It also allows the programmer to write functions to do conceptually different things depending on parameters, but that's a Real Bad Idea.)

5. c++ - Overloading of virtual functions - Stack Overflow

   stackoverflow.com/questions/42816834

   Here, since you're calling these functions on objects of static type Derived, C++ tries to find a function called fun in the Derived class. It finds your new function Derived::fun(float) , and due to the way C++ does name lookup in classes it does not look in the base class to find Base::fun(int) .

6. c++ - Overloading by return type - Stack Overflow

   stackoverflow.com/questions/9568852

   0. By norm c++ does not support return type overloading. However, it can be achieved by operator overloading though which is a advance techniques and not recommended for beginner. So this is achievable in following techniques: int getInt(){ int a = 65; return a;} char getChar(){ int a = 65; return (char) a; }

7. c++ issue with function overloading in an inherited class

   stackoverflow.com/questions/14212190

   All you need is a using: class classB:public classA{. public: using classA::func; void func(int){}; }; It doesn't search the base class for func because it already found one in the derived class. The using statement brings the other overload into the same scope so that it can participate in overload resolution.

8. c++ - function template overloading - Stack Overflow

   stackoverflow.com/questions/2174300

   I don't understand why some other answers mentions that template functions and function overloading doesn't mix. They certainly do, and there are special rules how the function to call is selected. 14.5.5. A function template can be overloaded with other function templates and with normal (non-template) functions.

9. c++ - Differentiate between function overloading and function ...

   stackoverflow.com/questions/11912022

   Overloading a method (or function) in C++ is the ability for functions of the same name to be defined as long as these methods have different signatures (different set of parameters). Method overriding is the ability of the inherited class rewriting the virtual method of the base class. a) In overloading, there is a relationship between methods ...

10. c++ - Difference between function overloading and method...

    stackoverflow.com/questions/9649005

    The quote snippet there is actually backward: method overloading is object-oriented, because otherwise you don't have objects with methods to be overloaded. Function is more generic, since it applies to free functions. In C++ we only have functions - free functions and member functions.

11. Function overloading in C++ - Stack Overflow

    stackoverflow.com/questions/8173492

    Integer types get automatically cast around a lot. So if you have a function overloaded on an int and a double, the compile will pick the int function if called with a constant that is an integer. If you didn't have the int version, the compiler would select the double one. And among various integer types, the compiler prefers int for integer ...