Re: [Boost-users] [format] compilation error

Subject: Re: [Boost-users] [format] compilation error
From: Anthony Foglia (AFoglia_at_[hidden])
Date: 2009-09-18 11:06:27

Johan Råde wrote:
> I'm using Boost 1.38 and Visual Studio 2008.
> The following code
> 
       [includes and usings omitted]
> 
>     ostream& operator<<(ostream& os, const vector<int>& v)
>     {
           [function body omitted]
>     }
> 
>     void f()
>     {
>         vector<int> v;
>         str(format("%1%") % v);
> 
>     }
> 
> fails to compile with the error message
> 
>     1>c:\libraries\boost\boost_1_38_0\boost\format\feed_args.hpp(100) : 
> error C2679: binary '<<' : no operator found which takes a right-hand 
> operand of type 'const std::vector<_Ty>' (or there is no acceptable 
> conversion)
>     1>        with
>     1>        [
>     1>            _Ty=int
>     1>        ]
> 
> Why? How do I fix this?

        The problem is that in feed_args.hpp when operator<< is called, the 
compiler looks in the namespaces boost and std.  The former because 
that's the current namespace, and std because that's the namespace of 
the vector argument.  It finds some candidates, none of which work for 
vector<int> and throws the error.  It never bothers to check the global 
namespace.

        I don't know if there's a recommended solution for this, but you can 
try specializing the operator<< in std namespace

namespace std {
template<>
ostream & operator<<(ostream & os, const vector<int>& v)
{
     // Your code here
}
}

        I believe that is allowed because you are fully specializing the 
template.  You aren't allowed to overload for a general vector<T> though.

        For a better explanation, check out this old GotW:
http://www.gotw.ca/publications/mill08.htm

-- 
Anthony Foglia
Princeton Consultants
[phone] x233