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Hello everybody,

How to avoid using struct IsDataType for comparison for find_if
algorithm but use something like that what is now commented out in the
code below?

Regards,
Misiu

template<typename T>
struct IsDataType
{
bool operator() (const boost::any& operand)
{
return (operand.type() ==
typeid(boost::shared_ptr<DataDistributor<T> >));
}
};

template<typename T>
static DataDistributor<T>* GetDataDistributor()
{
std::list<boost::any>::iterator iter =
std::find_if(m_distrubutors.begin(), m_distrubutors.end(),
IsDataType<T>());

//error C2248: 'type_info::type_info' : cannot access private member
//declared in class 'type_info' boost\tuple\detail\tuple_basic.hpp 477

// std::find_if(m_distrubutors.begin(), m_distrubutors.end(),
// boost::lambda::bind(&boost::any::type, _1) ==
// typeid(boost::shared_ptr<DataDistributor<T> >));
....
}

Misiu wrote:
> Hello everybody,
>
> How to avoid using struct IsDataType for comparison for find_if
> algorithm but use something like that what is now commented out in the
> code below?

You should ask on the boost mailing list.

"Misiu" <> wrote in message news:epq0qf$6l0$...
> Hello everybody,
>
> How to avoid using struct IsDataType for comparison for find_if algorithm
> but use something like that what is now commented out in the code below?
>
> Regards,
> Misiu
>
> template<typename T>
> struct IsDataType
> {
> bool operator() (const boost::any& operand)
> {
> return (operand.type() ==
> typeid(boost::shared_ptr<DataDistributor<T> >));
> }
> };
>
> template<typename T>
> static DataDistributor<T>* GetDataDistributor()
> {
> std::list<boost::any>::iterator iter =
> std::find_if(m_distrubutors.begin(), m_distrubutors.end(),
> IsDataType<T>());
>
> //error C2248: 'type_info::type_info' : cannot access private member
> //declared in class 'type_info' boost\tuple\detail\tuple_basic.hpp 477
> // std::find_if(m_distrubutors.begin(), m_distrubutors.end(),
> // boost::lambda::bind(&boost::any::type, _1) ==
> // typeid(boost::shared_ptr<DataDistributor<T> >));
> ...
> }

You can't copy-construct a type_info structure. You'll need boost::ref.

- Sylvester

>
> You can't copy-construct a type_info structure. You'll need boost::ref.
>
Thanks a lot.
Misiu