Re: static const integral members

From:
Carl Barron <cbarron413@adelphia.net>
Newsgroups:
comp.lang.c++.moderated
Date:
Sun, 13 Apr 2008 12:45:57 CST
Message-ID:
<130420080440144674%cbarron413@adelphia.net>
In article
<1d29a464-41dc-4986-97f6-549943e2f015@b64g2000hsa.googlegroups.com>,
Daniel Kr?gler <daniel.kruegler@googlemail.com> wrote:

On 12 Apr., 20:06, Carl Barron <cbarron...@adelphia.net> wrote:

In article
<c908380e-e078-4863-a31c-74ff14e9f...@x41g2000hsb.googlegroups.com>,

Daniel Kr?gler <daniel.krueg...@googlemail.com> wrote:

On 11 Apr., 19:34, "Roman.Perepeli...@gmail.com"
<Roman.Perepeli...@gmail.com> wrote:

I have a question regarding the following snippet of code.

struct foo
{
   static const int value = 1;
};

int main()
{
   int bar = foo::value;
}

Is it valid C++?


According to the wording of ISO 14882/2nd edition
this program violates the one definition rule,
because foo::value is not defined, but used, as
shown by the quotes you provided.


   Strange interpretation:
so the value const in std::integral_constant<typename T, T v>::value
is not defined either!.


I cannot follow your reasoning, mind to explain?

Just because [meta.help] does not *show* you a definition
for std::integral_constant<typename T, T v>::value is
not a proof, that it's definition is not needed.

A typical implementation looks like this (modulo namespace):

template <class T, T v>
struct integral_constant {
   static const T value = v;
   typedef T value_type;
   typedef integral_constant<T,v> type;
};

template<class T, T v> const T integral_constant<T, v>::value;

For similar reasons you could argue that no definitions
of std::basic_string<..>::npos are required. But how can
the library implementor know, whether the user-code
takes e.g. a pointer or reference to this constant?
In this cases a definition of npos will be required
in both the current and the new C++ standard.

that is

// std headers not listed here

int main()
{
    bool t = std::is_integral<double>::value; // is not define either!!!'
    std::cout << t << '\n'; // t == 0 is it not??
    //....

}

I seem to think there is something definitely wrong with this strange
interpretation. :)


As described above, the library implementation must
already provide a definition of std::is_integral<double>::value.
By including <type_traits> you have ensured that your
program contains the definition. Since std::is_integral
is a template, the compiler has to ensure that
the ODR is fulfilled, even if the program contains
multiple translation units that include <type_traits>,
see [basic.def.odr]/5:

"There can be more than one definition of a [..] static data
member of a class template (14.5.1.3), [..] in a program
provided that each definition appears in a different translation
unit, and provided the definitions satisfy the following
requirements[..]"

Greetings from Bremen,

Daniel Kr?gler

  The point is
   struct foo
   {
      static const int value = 1;
   };

   does not automatically violate ODR only different defintions of foo
would do so. The sniplet does not involve multiple defintions of foo
or
foo::value. and the sniplet is legal and valid.

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