why can't we assign weaker privilege in subclass

Question

I have a class which has a method whose access specifier by default is public. Now, I would like to extend this class in a subclass and I want to override this method to hav

Answer 1

I think the short answer is that the compiler writers have set the rules to work this way. LSP has nothing to do with the problem at hand.

The only reason I can think of to have this restriction is that when a subclass derives from an interface, as a client programmer, you expect to be able to call all the accessible methods of the interface from a reference to the derived class.

Just suppose that you could write the code that the OP has shown. If you have a reference to the derived class you should be able to call any public members of the derived class (though there are none in this case). However, pass the reference as a parameter to a method which takes a reference to the base class and the method will expect to call any public or package method, which is foo. This is the LSP that other contributors are looking for!

C++ example:

class Superclass{
public:
virtual void foo(){ cout << "Superclass.foo" << endl; }
};

class Subclass: public Superclass{
virtual void foo(){ cout << "Subclass.foo" << endl; }
};

int main(){
Superclass s1;
s1.foo() // Prints Superclass.foo

Subclass s2;
// s2.foo();  // Error, would not compile

Superclass& s1a=s2; // Reference to Superclass 'pointing' to Subclass

s1a.foo(); // Compiles OK, Prints Subclass.foo()
}