Re: Why can't a pure virtual function have an inline definition?

From:
David Krauss <potswa@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 16 Sep 2010 15:53:59 CST
Message-ID:
<867a8a63-990a-4eb5-badc-9de65721f9ba@c13g2000vbr.googlegroups.com>
On Sep 15, 11:27 am, Frank Buss <f...@frank-buss.de> wrote:

armen.tsirunyan wrote:

It is clearly stated in the current C++ International Standard (2003)
that no virtual function declaration shall contain both a pure
specifier and a definition. I wonder what is the reason of prohibiting
this? I mean, the feature is obviously trivial to implement, it is
more uniform and, well, quite useful. So what considerations am I
missing? Thank you.


It's just an unfortunate syntax quirk. The "= 0" construct has not
been assigned a slot relative to the exception-specification, ref-
qualifier, cv-qualifier, attribute-specifier, and trailing-return-
type.

This is the same as in the ISO IEC 14882-1999, see chapter 10.4.2, and it
makes sense, because the idea of a "pure virtual" function is, that there
is no definition, so why do you think it would be useful? Providing a
defintion for a virtual function might be useful, but I don't see how it
would make sense for pure virtual functions.


"Pure virtual" means that a function is not inserted into the virtual
lookup system, i.e. it's vtable entry is set to 0 or NULL. It may
still be defined, but to call it requires (perhaps ironically)
qualified my_class::my_purevirt_fn() syntax.

But it could be useful to declare an abstract class without any pure
virtual function, like it is possible in Java. I don't know if C++0x
supports this.


Just declare the destructor pure virtual. You can still provide a
definition for that destructor and it will be called not-in-charge.

#include <iostream>

struct S {
   virtual ~S() = 0;
};

S::~S() { std::cout << __func__ << '\n'; }

struct T : S {
   ~T() { std::cout << __func__ << '\n'; }
};

int main() {
   T t;
}

  output:
~T
~S

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