Sunday, November 30, 2014

C++ Specific Questions - Overloading VS Overriding

Ref: http://stackoverflow.com/questions/429125/override-and-overload-in-c

Overloading generally means that you have two or more functions in the same scope having the same name. The function that better matches the arguments when a call is made wins and is called. Important to note, as opposed to calling a virtual function, is that the function that's called is selected at compile time. It all depends on the static type of the argument. If you have an overload for B and one for D, and the argument is a reference to B, but it really points to a D object, then the overload for B is chosen in C++. That's called static dispatch as opposed to dynamic dispatch. You overload if you want to do the same as another function having the same name, but you want to do that for another argument type. Example:
void print(Foo const& f) {
    // print a foo
}

void print(Bar const& bar) {
    // print a bar
}
they both print their argument, so they are overloaded. But the first prints a foo, and the second prints a bar. If you have two functions that do different things, it's considered bad style when they have the same name, because that can lead to confusion about what will happen actually when calling the functions. Another usecase for overloading is when you have additional parameters for functions, but they just forward control to other functions:
void print(Foo & f, PrintAttributes b) { 
    /* ... */ 
}

void print(Foo & f, std::string const& header, bool printBold) {
    print(f, PrintAttributes(header, printBold));
}
That can be convenient for the caller, if the options that the overloads take are often used.
Overriding is something completely different. It doesn't compete with overloading. It means that if you have a virtual function in a base class, you can write a function with the same signature in the derived class. The function in the derived class overrides the function of the base. Sample:
struct base {
    virtual void print() { cout << "base!"; }
}

struct derived: base {
    virtual void print() { cout << "derived!"; }
}
Now, if you have an object and call the print member function, the print function of the derived is always called, because it overrides the one of the base. If the function print wasn't virtual, then the function in the derived wouldn't override the base function, but would merely hide it. Overriding can be useful if you have a function that accepts a base class, and every one that's derived from it:
void doit(base &b) {
    // and sometimes, we want to print it
    b.print();
}
Now, even though at compile time the compiler only knows that b is at least base, print of the derived class will be called. That's the point of virtual functions. Without them, the print function of the base would be called, and the one in the derived class wouldn't override it.



You overload functions for three reasons:
  1. To provide two (or more) functions that perform similar, closely related things, differentiated by the types and/or number of arguments it accepts. Contrived example:
    void Log(std::string msg); // logs a message to standard out
    void Log(std::string msg, std::ofstream); // logs a message to a file
  2. To provide two (or more) ways to perform the same action. Contrived example:
    void Plot(Point pt); // plots a point at (pt.x, pt.y)
    void Plot(int x, int y); // plots a point at (x, y)
  3. To provide the ability to perform an equivalent action given two (or more) different input types. Contrived example:
    wchar_t      ToUnicode(char c);
    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.

Overriding a function is entirely different, and serves an entirely different purpose. Function overriding is how polymorphism works in C++. You override a function to change the behavior of that function in a derived class. In this way, a base class provides interface, and the derived class provides implementation.

1 comment:

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