Inheritance and casting with templates not working as expected

Question

I've been playing around with template and inheritance but there is something strange about using virtual function members with template parameters when performing a cast to the base class. They seem to work using "direct inheritance" but not if I "defer" the inheritance later on.

A bit of code to illustrate:

Example [1]

struct CastExBase
  {
  virtual void f() {}
  };

template<class RT>
struct CastExA : CastExBase
  {
  void f() {std::cout << "CastExA" << std::endl;}
  virtual void g() {std::cout << "g - A" << std::endl;}
  virtual RT h() {std::cout << "h - A" << std::endl;}
  };

struct CastExB
  {
  void execF() {std::cout << "CastExB" << std::endl;}
  void g() {std::cout << "g - B" << std::endl;}
  int h() {std::cout << "h - B" << std::endl;}
  };

struct CastExC :
    public CastExA<int>,
    protected CastExB
  {
  void f() override
    {
    (static_cast<CastExB*>(this))->execF();
    }

  void g() override
    {
    (static_cast<CastExB*>(this))->g();
    }
  };

Test case:

  CastExBase* a2 = new CastExC();
  CastExA<int>* a3 = (CastExA<int>*) a2;
  a3->g(); // This prints g - B as expected
  a3->h(); // This prints h - A ... why???

Why a3->h() does not print h - B?

I tried also another test directly inheriting from the base class and in this case it correctly works.

Example [2]

struct CastExDBase
      {
      };

    template<class T>
    struct CastExD : CastExDBase
      {
      virtual T f() {std::cout << "CastExD" << std::endl;}
      };

    struct CastExE : CastExD<int>
      {
      int f() {std::cout << "CastExE" << std::endl;}
      };

Test case:

  CastExDBase* d1 = new CastExE();
  CastExD<int>* d2 = (CastExD<int>*) d1;
  d2->f(); // This prints CastExE as expected

Is this related to UB?

`CastExA* a3 = (CastExA*) a2;` - You cannot downcast with a C-style cast, you need to `dynamic_cast` here. — Holt, Jan 16 '19 at 08:42
ideally yes but I know that the type will be CastExA* so should be ok to use it. Anyway, I tried also to dynamic cast it but same result — svoltron, Jan 16 '19 at 08:47
General tip: Never do C-style casts in C++. If you ever need to do it it's usually a red flag that you're doing something wrong. — Some programmer dude, Jan 16 '19 at 08:47
Closely related, possibly dupe: https://stackoverflow.com/questions/19528338/how-to-override-a-function-in-another-base-class — Holt, Jan 16 '19 at 08:59

score 5 · Accepted Answer · answered Jan 16 '19 at 08:57

Even if CastExC inherits both CastExA<int> and CstExB, the definition of h() in CstExB will not override the "definition" of h() in CastExA<int> because CastExA<int> and CstExB are not related. If you try to do this:

CastExBase* a2 = new CastExC();
CastExC* a3 = (CastExC*) a2;
a3->h();

You will get an ambiguous request for h(). If you want to overload, you need to do this manually:

struct CastExC: protected CastExB, public CastExA<int> {
    virtual int h() override { return CastExB::h(); }
};

score 4 · Answer 2 · edited Jan 16 '19 at 09:00

4

There is no h() in CastExC. Maybe that's why.

Note: CastExA and CastExB are not related.

edited Jan 16 '19 at 09:00

ETO

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answered Jan 16 '19 at 08:52

Marek Jamiołkowski

41
4

Damn it was indeed that... as always the simplest solution... thank you – svoltron Jan 16 '19 at 08:58

Inheritance and casting with templates not working as expected

2 Answers2