overloading

What is the difference between dynamic and static polymorphism in Java?

Polymorphism 1. Static binding/Compile-Time binding/Early binding/Method overloading.(in same class) 2. Dynamic binding/Run-Time binding/Late binding/Method overriding.(in different classes) overloading example: class Calculation { void sum(int a,int b){System.out.println(a+b);} void sum(int a,int b,int c){System.out.println(a+b+c);} public static void main(String args[]) { Calculation obj=new Calculation(); obj.sum(10,10,10); // 30 obj.sum(20,20); //40 } } overriding example: class Animal { public void move(){ … Read more

PHP function overloading

You cannot overload PHP functions. Function signatures are based only on their names and do not include argument lists, so you cannot have two functions with the same name. Class method overloading is different in PHP than in many other languages. PHP uses the same word but it describes a different pattern. You can, however, … Read more

How do I use method overloading in Python?

It’s method overloading, not method overriding. And in Python, you historically do it all in one function: class A: def stackoverflow(self, i=’some_default_value’): print ‘only method’ ob=A() ob.stackoverflow(2) ob.stackoverflow() See the Default Argument Values section of the Python tutorial. See “Least Astonishment” and the Mutable Default Argument for a common mistake to avoid. See PEP 443 … Read more

TypeScript function overloading

When you overload in TypeScript, you only have one implementation with multiple signatures. class Foo { myMethod(a: string); myMethod(a: number); myMethod(a: number, b: string); myMethod(a: any, b?: string) { alert(a.toString()); } } Only the three overloads are recognized by TypeScript as possible signatures for a method call, not the actual implementation. In your case, I … Read more

Template Specialization VS Function Overloading

Short story: overload when you can, specialise when you need to. Long story: C++ treats specialisation and overloads very differently. This is best explained with an example. template <typename T> void foo(T); template <typename T> void foo(T*); // overload of foo(T) template <> void foo<int>(int*); // specialisation of foo(T*) foo(new int); // calls foo<int>(int*); Now … Read more

Why do multiple-inherited functions with same name but different signatures not get treated as overloaded functions?

Member lookup rules are defined in Section 10.2/2 The following steps define the result of name lookup in a class scope, C. First, every declaration for the name in the class and in each of its base class sub-objects is considered. A member name f in one sub-object B hides a member name f in … Read more