Re: A problem about virtual function?

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Tue, 16 Oct 2007 14:42:25 GMT
On 2007-10-16 16:09, Wayne Shu wrote:

Hi everyone,

    Today I meet a problem about virtual function. Consider the
following little program.

#include <iostream>

class base
    int i;
    virtual void print() { std::cout << "base" << std::endl; }
    virtual ~base() {}

class derived : public base
    virtual void print() { std::cout << "derived" << std::endl; }

int main()
    base b;

    new (&b) derived;

    (&b)->print(); // (1)
    base *p = &b;

    return 0;

I have try the VC2005 Express Edition and GCC 3.4.2
and the result is :

Why (1) statement print "base"?

First, I am not sure if it is legal to call the destructor of an
automatic object and then construct a new object of a type derived from
the first object's type in the same place, it seems very suspect to me.

My *guess* of what is happening is this. First you create an object of
type base with automatic storage (on the stack). An object can never
have polymorphic behaviour (only a pointer/reference to an object), an
object always have a known type (even though it might not be known when
the program is compiled). More importantly, all objects on the stack
have a type which is know at compile-time.

So when you do (&b)->print() the compiler knows that the type of b is
base, and thus the only possible outcome of that line is a call to
base::print(). It is an optimisation, the compiler will try to not make
a virtual call if it does not have to since they involve an extra

On the other hand when you create a pointer to the object and then call
print() on that all the compiler knows is that you call print() on a
pointer to base, the actual type of the object that the pointer points
to is not know and a virtual call must be made.

Erik Wikstr??m

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