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Home/ Questions/Q 8654915
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T14:54:12+00:00 2026-06-12T14:54:12+00:00

I have the following method in which I’m using boost::variant. I try to get

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I have the following method in which I’m using boost::variant. I try to get the value, based on type T. If boost::get<T> fails I want to handle that in a special way if T is an int or unsigned int. Is there any way to know if T is an int or unsigned int?

I don’t think I can use template specialization in this case, can I?

EDIT: Also, I don’t yet have access to C++11 (soon I hope)

template < typename T, typename C, void (C::*setterFcn)(const T&) >
void binder( const Variant& value_var, C* c )
{
    const T* typeData = boost::get<T>(&value_var);

    if ( NULL == typeData )
    {
        // Need to check for int or unsigned int here somehow
    }

    (((C*) c)->*(setterFcn))(*typeData);
}
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  1. Editorial Team
    Editorial Team
    2026-06-12T14:54:13+00:00Added an answer on June 12, 2026 at 2:54 pm

    In C++11 you can use std::is_same and in C++03 you can do something like this:

    template <typename T1, typename T2>
    class is_same
    {
    public: 
        static bool const value = false;
    };
    
    template <typename T>
    class is_same<T, T>
    {
    public: 
        static bool const value = true;
    };
    

    and use it exactly as C++11 standard version.

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