c++ - Meaning of 'const' last in a function declaration of a class?

When you add the const keyword to a method the this pointer will essentially become a pointer to const object, and you cannot therefore change any member data. (Unless you use mutable, more on that later).

The const keyword is part of the functions signature which means that you can implement two similar methods, one which is called when the object is const, and one that isn't.

#include 
class MyClass
{
private:
    int counter;
public:
    void Foo()
    {
        std::cout << "Foo" << std::endl;
    }
    void Foo() const
    {
        std::cout << "Foo const" << std::endl;
    }
};
int main()
{
    MyClass cc;
    const MyClass& ccc = cc;
    cc.Foo();
    ccc.Foo();
}

This will output

Foo
Foo const

In the non-const method you can change the instance members, which you cannot do in the const version. If you change the method declaration in the above example to the code below you will get some errors.

    void Foo()
    {
        counter++; //this works
        std::cout << "Foo" << std::endl;
    }
    void Foo() const
    {
        counter++; //this will not compile
        std::cout << "Foo const" << std::endl;
    }

This is not completely true, because you can mark a member as mutable and a const method can then change it. It's mostly used for internal counters and stuff. The solution for that would be the below code.

#include 
class MyClass
{
private:
    mutable int counter;
public:
    MyClass() : counter(0) {}
    void Foo()
    {
        counter++;
        std::cout << "Foo" << std::endl;
    }
    void Foo() const
    {
        counter++;
        std::cout << "Foo const" << std::endl;
    }
    int GetInvocations() const
    {
        return counter;
    }
};
int main(void)
{
    MyClass cc;
    const MyClass& ccc = cc;
    cc.Foo();
    ccc.Foo();
    std::cout << "Foo has been invoked " << ccc.GetInvocations() << " times" << endl;
}

which would output

Foo
Foo const
Foo has been invoked 2 times