Wrong usage of static_cast: is output undefined or predicted?

[Alex Vinokur]

Here is some program in which static_cast is (wrongly) used instead of dynamic_cast.

Is output of the program undefined or predicted?

=========== C++ code : foo.cpp : BEGIN ===========
#include <iostream>
using namespace std;

struct B
{
virtual void foo1 () { cout << "B::foo1" << endl; }
virtual void foo2 () { cout << "B::foo2" << endl; }
};
struct D1 : public B
{
void foo1 () { cout << "D1::foo1" << endl; }
};
struct D2 : public B
{
void foo2 () { cout << "D2::foo2" << endl; }
};

int main ()
{
B* pB = new D1;
// --------------------------
// Of course, here we should use dynamic_cast
D2* pD2 = static_cast<D2*> (pB);
// --------------------------

pD2->foo1();
pD2->foo2();

return 0;
}

=========== C++ code : foo.cpp : END =============



=========== Compilation & Run : BEGIN =============

$ g++ --version
g++ (GCC) 3.3.3 (cygwin special)
[---omitted---]

$ g++ -W -Wall foo.cpp
// No errors, no warnings

$ a
D1::foo1
B::foo2


=========== Compilation & Run : END ===============

 

[Alf P. Steinbach]

* Alex Vinokur:

Here is some program in which static_cast is (wrongly) used instead of dynamic_cast.

Is output of the program undefined or predicted?

UB.

The code will probably work on most compilers since the derived classes
do not add anything to or otherwise change the object memory layout.

// --------------------------
// Of course, here we should use dynamic_cast

Just say no to casting.