Inheitance not working while workig with pointer

Question

although we are printing bval, output is showing value of dval as well. Whats actual logic does compiler using? My expected output was 00000 00000, but I m getting output as 00000 01010.

#include <iostream>
using namespace std;

class base {
 public:
   int bval;
   base() { bval = 0; }
};

class deri : public base {
  public:
    int dval;
    deri() { dval = 1; }
 };

 void SomeFunc(base *arr , int size) {
   for(int i = 0; i < size; i++, arr++)
     cout << arr-> bval;
   cout<<endl;
 }

 int main() {
   base BaseArr[5];
   SomeFunc(BaseArr, 5);
   deri DeriArr[5];
   SomeFunc(DeriArr, 5);
   return 0;
 }

Undefined behavior. You cannot treat arrays polymorphically. Searching for duplicate... — CB Bailey, Apr 24 '13 at 10:44
or... http://stackoverflow.com/questions/1411844/polymorphism-pointers-to-arrays — CB Bailey, Apr 24 '13 at 10:47

score 3 · Answer 1 · answered Apr 24 '13 at 10:47

You are cheating the system, and getting "slicing".

Since deri is larger than base, when you do arr++, the pointer is only advanced an base object forward.

The solution is to have an array of pointers, and passing a base **arr, like this:

base *foo[5];
base BaseArr[5];
SomeFunc(BaseArr,5);
for(int i = 0; i < 5; i++)
{ 
   foo[i] = &BaseArr[i];
}
SomeFunc(foo, 5);

Inheitance not working while workig with pointer

1 Answers1