Unnamed/anonymous namespaces vs. static functions

Question

A feature of C++ is the ability to create unnamed (anonymous) namespaces, like so:

namespace {
    int cannotAccessOutsideThisFile() { ... }
} // namespace

You would think that such a feature would be useless -- since you can't specify the name of the namespace, it's impossible to access anything within it from outside. But these unnamed namespaces are accessible within the file they're created in, as if you had an implicit using-clause to them.

My question is, why or when would this be preferable to using static functions? Or are they essentially two ways of doing the exact same thing?

Answer 1

From experience I'll just note that while it is the C++ way to put formerly-static functions into the anonymous namespace, older compilers can sometimes have problems with this. I currently work with a few compilers for our target platforms, and the more modern Linux compiler is fine with placing functions into the anonymous namespace.

But an older compiler running on Solaris, which we are wed to until an unspecified future release, will sometimes accept it, and other times flag it as an error. The error is not what worries me, it's what it might be doing when it accepts it. So until we go modern across the board, we are still using static (usually class-scoped) functions where we'd prefer the anonymous namespace.

Answer 2

There is one edge case where static has a surprising affect (at least it was to me). The C++03 Standard states in 14.6.4.2/1:

For a function call that depends on a template parameter, if the function name is an unqualified-id but not a template-id, the candidate functions are found using the usual lookup rules (3.4.1, 3.4.2) except that:

• For the part of the lookup using unqualified name lookup (3.4.1), only function declarations with external linkage from the template definition context are found.
• For the part of the lookup using associated namespaces (3.4.2), only function declarations with external linkage found in either the template definition context or the template instantiation context are found.

...

The below code will call foo(void*) and not foo(S const &) as you might expect.

template <typename T>
int b1 (T const & t)
{
  foo(t);
}

namespace NS
{
  namespace
  {
    struct S
    {
    public:
      operator void * () const;
    };

    void foo (void*);
    static void foo (S const &);   // Not considered 14.6.4.2(b1)
  }

}

void b2()
{
  NS::S s;
  b1 (s);
}

In itself this is probably not that big a deal, but it does highlight that for a fully compliant C++ compiler (ie. one with support for export) the static keyword will still have functionality that is not available in any other way.

// bar.h
export template <typename T>
int b1 (T const & t);

// bar.cc
#include "bar.h"
template <typename T>
int b1 (T const & t)
{
  foo(t);
}

// foo.cc
#include "bar.h"
namespace NS
{
  namespace
  {
    struct S
    {
    };

    void foo (S const & s);  // Will be found by different TU 'bar.cc'
  }
}

void b2()
{
  NS::S s;
  b1 (s);
}

The only way to ensure that the function in our unnamed namespace will not be found in templates using ADL is to make it static.

Update for Modern C++

As of C++ '11, members of an unnamed namespace have internal linkage implicitly (3.5/4):

An unnamed namespace or a namespace declared directly or indirectly within an unnamed namespace has internal linkage.

But at the same time, 14.6.4.2/1 was updated to remove mention of linkage (this taken from C++ '14):

For a function call where the postfix-expression is a dependent name, the candidate functions are found using the usual lookup rules (3.4.1, 3.4.2) except that:

• For the part of the lookup using unqualified name lookup (3.4.1), only function declarations from the template definition context are found.

• For the part of the lookup using associated namespaces (3.4.2), only function declarations found in either the template definition context or the template instantiation context are found.

The result is that this particular difference between static and unnamed namespace members no longer exists.

Answer 3

In addition if one uses static keyword on a variable like this example:

namespace {
   static int flag;
}

It would not be seen in the mapping file

Answer 4

Personally I prefer static functions over nameless namespaces for the following reasons:

• it's obvious and clear from function definition alone that it's private to the translation unit where it's compiled. With namesless namespace you might need to scroll and search to see if a function is in a namespace.

• functions in namespaces might be treated as extern by some (older) compilers. In VS2017 they are still extern. For this reason even if a function is in nameless namespace you might still want to mark them static.

• static functions behave very similar in C or C++, while nameless namespaces are obviously C++ only. nameless namespaces also add extra level if indentation and I don't like that :)

So, I'm happy to see that use of static for functions isn't deprecated anymore.