C++ delete a class pointer & polymorphism - c++

Question

I am learning about polymorphism, and this is something I have stumbled on. I cant find satisfying answer and testing the following code doesnt yield expected result.
#include <stdio.h>
#include <iostream>
class Base {
public:
Base(){
std::cout<<"Constructing Base" << std::endl;
};
virtual int get_a(){
return p_a;
};
virtual int check(){
std::cout<<"check Base" << std::endl; return 2;
};
virtual ~Base(){
std::cout<<"Destroying Base" << std::endl;
};
int p_a = 4;
};
class Derive: public Base {
public:
Derive(){
std::cout<<"Constructing Derive" << std::endl;
};
int get_a(){
return p_a;
};
int check(){
std::cout<<"check Derive" << std::endl;
return 3;
};
~Derive(){ std::cout<<"Destroying Derive" << std::endl; };
int p_a = 2;
};
int main() {
Base *basePtr = new Derive();
delete basePtr;
basePtr->check();
std::cout << "p_a: " << basePtr->get_a() << std::endl;
return 1;
}
Console output :
Constructing Base // line 1
Constructing Derive // line 2
Destroying Derive // line 3
Destroying Base // line 4
check Base // line 5
p_a: 4 // line 6
I see why I get line 1-4, the basePtr is pointer to Derive, which inherits from Base, which implements virtual functions.
My 1. expectation : After calling delete, the pointer basePtr should not be able to deliver the function call ->check(), and also there should be no value p_a.
My 2. expectation : I would expect the value p_a from Derive (p_a = 2), to appear on the output, because basePtr stores the pointer of Derive.
Could someone correct my thinking ?

Answer 1

This is undefined behavior. Whatever results you get, are invalid.
delete basePtr;
This destroys the object.
basePtr->check();
After destroying the object, the shown code attempts to dereference a pointer to the destroyed instance of the class, and invoke a method of the destroyed object. Invoking a method of a destroyed object is undefined behavior.