Re: What Standard says on declared only symbols

=?iso-8859-1?q?Daniel_Kr=FCgler?= <>
Wed, 24 Oct 2007 12:19:34 CST
On 23 Okt., 21:04, Adam Badura <> wrote:

    Lets assume I have declared a symbol (lets say, a function "void
f()") but never defined it as well as never used it. All compilers
that I have worked with do not make any problems, they just ignore
"f". But do Standard makes any guarantee on that? Or is such program
not well formed?

The standard distinguishes several form of "use" in the section
about the one-definition-rule (ODR). The most relevant parts related
to your issue are probably the following two:


"[..] A virtual member function is used if it is not pure.[..]"

"Every program shall contain exactly one definition of every non-inline
function or object that is used in that program; no diagnostic required.
The definition can appear explicitly in the program, it can be found in
the standard or a user-defined library, or (when appropriate) it is
implicitly defined (see 12.1, 12.4 and 12.8).[..]"

Basically this says that every virtual function (which is not pure)
has to be viewed as used and thus needs to be defined.

Besides the definition issue you are also stumbling across
an instantiation issue, which is described here:


"An implementation shall not implicitly instantiate a function template,
a member template, a non-virtual member function, a member class or a
static data member of a class template that does not require instantiation.
It is unspecified whether or not an implementation implicitly instantiates
a virtual member function of a class template if the virtual member
function would not otherwise be instantiated.[..]"

    This question arose from a simple problem. I had a template
container (TreeInfo). I wanted to make it serializable (by inheriting
from my own IWritable interface). The interface introduced two (pure
virtual) functions "load" and "save". I wrote them for the TreeInfo
assuming the contained type was serializable as well. In case the
contained type was not serializable the code was incorrect (however
syntax was OK). But I thought that if the contained type was not
serializable then no one will ever serialize the container and thous
"load" and "save" would not be used and thous no error would occur
during compilation. However it didn't work this way. (Maybe it is
compiler specific, this time I used MSVC 8.0.) It seems that since
"load" and "save" are virtual they must be present in vtable and thous
must be compiled even if not used.

Yes, this is the practical explanation and for this reason above
quoted standard phrase concerning instantiation freedom of
virtual function applies.

    I try to solve the problem somehow. I can make just a
TreeInfoSerializable which will inherit TreeInfo as well as IWritable
and expect my users to use either one as needed. I could try some
tricks with template metaprogramming (this seems most elegant however
is likely to be worst because the code is in a large project and this
could cause need for a lot of changes). I thought also on some
implementation tricks they would however relay on defined but not
declared functions.

I don't understand this last sentence, because every definition
is also a declaration - would you explain?

I don't know, what you mean with above suggestion concerning
template metaprogramming, but a possible solution is to use
SFINAE here:

struct IWritable {
  virtual void write() = 0;
  virtual void read() = 0;

template <class T>
struct is_writable {
  enum {value = std::is_base_of<IWritable, T>::value };

template <typename T, class Enable = void>
struct TreeInfo;

template <typename T>
struct TreeInfo<T, typename
  std::enable_if<!is_writable<T>::value>::type> {
  T d;

template <typename T>
struct TreeInfo<T, typename
  std::enable_if<is_writable<T>::value>::type> : IWritable {
  void write(){
  void read(){;
  T d;

struct Writable : IWritable {
  void write(){}
  void read(){}

TreeInfo<int> it; // OK, uses the non-writable TreeInfo
TreeInfo<Writable> wt; // OK, uses writable TreeInfo

HTH and Greetings from Bremen,

Daniel Kr?gler

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