using const causes compiler errors

[skew]

I have the following class

class Test
{
public:
	void DoSomeStuff()
	{
		cout << "Doing some stuff ..." << endl;
	}
};

Using the class like follows, I do not get a compiler error. All works fine

Test test;
test.DoSomeStuff();

But if I write const before "Test test" like so

const Test test;
test.DoSomeStuff(); // Compiler Error! 

I get two error messages from the compiler:

error C2662: 'Test::DoSomeStuff' : cannot convert 'this' pointer from 'const Test' to 'Test &'

IntelliSense: the object has type qualifiers that are not compatible with the member function

Why it doesn't work with const?

[Peter87]

You can only call the member functions that are marked with const on const objects.
Change line 4 to void DoSomeStuff() const. Now inside this function you will not be able to change member variables or call other non-const member functions.

[Wazzak]

An instantiation of an object that's declared constant can only invoke constant methods of the same class. For example:

class Test
{
    public:
        void DoSomeStuff() // Not constant
        {
            std::cout << "Test::DoSomeStuff()" << std::endl;
        }

        void DoSomeStuff() const // Constant
        {
            std::cout << "Test::DoSomeStuff() const" << std::endl;
        }
};

int main()
{
    const Test NewTest;
    NewTest.DoSomeStuff();
}

Here, the first definition of Test::DoSomeStuff() isn't constant because no const follows the function header. However, the second definition of Test::DoSomeStuff() is constant because const follows the function header.

It's quite simple: If const follows the function header, the function is considered read-only; that is, the function cannot alter any members of the class, regardless. A read-only function can only appear in a class.

Note this:
You can only invoke read-only methods from constant instantiations of the same class.

Compile the above code and see what happens.

_Wazzak

[skew]

Thanks!!