What is the 'override' keyword in C++ used for? [duplicate]

The override keyword serves two purposes:

1. It shows the reader of the code that "this is a virtual method, that is overriding a virtual method of the base class."
2. The compiler also knows that it's an override, so it can "check" that you are not altering/adding new methods that you think are overrides.

To explain the latter:

class base
{
  public:
    virtual int foo(float x) = 0; 
};


class derived: public base
{
   public:
     int foo(float x) override { ... } // OK
}

class derived2: public base
{
   public:
     int foo(int x) override { ... } // ERROR
};

In derived2 the compiler will issue an error for "changing the type". Without override, at most the compiler would give a warning for "you are hiding virtual method by same name".

And as an addendum to all answers, FYI: override is not a keyword, but a special kind of identifier! It has meaning only in the context of declaring/defining virtual functions, in other contexts it's just an ordinary identifier. For details read 2.11.2 of The Standard.

#include <iostream>

struct base
{
    virtual void foo() = 0;
};

struct derived : base
{
    virtual void foo() override
    {
        std::cout << __PRETTY_FUNCTION__ << std::endl;
    }
};

int main()
{
    base* override = new derived();
    override->foo();
    return 0;
}

Output:

zaufi@gentop /work/tests $ g++ -std=c++11 -o override-test override-test.cc
zaufi@gentop /work/tests $ ./override-test
virtual void derived::foo()

override is a C++11 keyword which means that a method is an "override" from a method from a base class. Consider this example:

   class Foo
   {
   public:
        virtual void func1();
   }

   class Bar : public Foo
   {
   public:
        void func1() override;
   }

If B::func1() signature doesn't equal A::func1() signature a compilation error will be generated because B::func1() does not override A::func1(), it will define a new method called func1() instead.