"a struct has public inheritance by default"

Question

"a struct has public inheritance by default" what does this statement really mean? And why is the following code in error just because I have omitted the keyword 'public' while deriving the class d from c??

struct c 
{
protected:
    int i;
public:
    c(int ii=0):i(ii){} 
    virtual c *fun();
};

c* c::fun(){
    cout<<"in c";
    return &c();
}

class d : c
{  
public:
    d(){}
    d* fun()
    {
        i = 9;
        cout<<"in d"<<'\t'<<i;
        return &d();
    }
};


int main()
{
    c *cc;
    d dd;
    cc = &dd;
    cc->fun();
}

Answer 1

It means that

struct c;
struct d : c

is equivalent to

struct d : public c

Your code is a class extending a struct:

struct c;
class d : c;

is equivalent to

class d : private c;

because class has private inheritance by default.

And it means that all inherited and not overriden/overloaded/hidden methods from c are private in d.

Answer 2

"a struct has public inheritance by default" means that this

struct Derived : Base {};

is equivalent to

struct Derived : public Base {};

Classes have private inheritance by default, so when you remove the public from a class inheritance you have the equivalent of

class Derived : private Base {};

In this private inheritance scenario, Derived does not have an is-a relationship with Base, it essentially has-a Base. So the conversion you are trying to attempt here:

cc = ⅆ

is not allowed.