How does RTTI work?

Question

I have some confusion regarding the RTTI mechanism in C++.

Suppose in have class A and class B that inherits from A. Now consider the following code:

B* b = new B();
A* a = dynamic_cast<A*>(b);

I know that polymorphic classes with virtual methods have virtual tables and vptr's, but I thought that the pointers only give information about the virtual functions. How does the program know at runtime the type of b, using vptr's and vtables?

Krakow

"I thought that the pointers only give information about the virtual functions" - why did you think that? The vptr points to all the dynamic type information - the vtable, and whatever is needed by RTTI. – Cristoforo

@MikeSeymour besides the vtable, what else is pointed at? – Krakow

Whatever implementation-specific metadata is needed to support RTTI. I'm afraid I don't know how it's generally implemented, but I'm sure Google could tell you if you're interested. – Cristoforo

This may or may not be helpful. – Transmundane

Answer 1

Imagine you have

struct B {
    virtual doSth() {
        cout << "hello";
    }
};
struct A : public B {
    doSth() {
        cout << "hello world";
    }
};

Now suppose A::doSth() is at 0x0f43 and B::doSth() is at 0x0a41

then dynamic_cast(b) can be implemented as (pseudo-code)

if ( (address pointed to by b::doSth()) == 0x0f43 ) {
    // cast is OK
} else { // (address pointed to by b::doSth()) == 0x0a41
    // not possible to cast
}

So you really just need b to hold a pointer to the right doSth() method to know its true type

Stipulation

I assume there is an implied line that says something like: void *b = (void *)new B() in there. If so, wouldn't it be possible that b is some totally unrelated thing, like an array of bytes, that happens to have a 0x0f43 at the right offset? – Discernible

If b is an array of bytes, the dynamic_cast won't even compile because b is not of type B. – Shanel

@Moby Dick. No it does not compile (or should not). See https://mcmap.net/q/830260/-dynamic_cast-from-quot-void-quot – Shanel