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Home/ Questions/Q 114099
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Editorial Team
Asked: May 11, 20262026-05-11T02:48:24+00:00 2026-05-11T02:48:24+00:00

I tried to compile the example posted ( C++ Service Providers ) and failed

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I tried to compile the example posted (C++ Service Providers) and failed to VS8 VC9. I have little experience with template.
Any suggestions?
Tanks.

These are the errors :
dictionarystl.cpp(40) : error C2663: ‘std::_Tree<_Traits>::find’ : 2 overloads have no legal conversion for ‘this’ pointer
dictionarystl.cpp(48) : error C2679: binary ‘[‘ : no operator found which takes a right-hand operand of type ‘const type_info *__w64 ‘ (or there is no acceptable conversion)

#include <typeinfo> #include <map> #include <string> using namespace std;  class SomeClass { public:     virtual ~SomeClass() {} // virtual function to get a v-table };  struct type_info_less {     bool operator() (const std::type_info* lhs, const std::type_info* rhs) const     {         return lhs->before(*rhs) != 0;     } };  class TypeMap {     typedef map <type_info *, void *, type_info_less> TypenameToObject;     TypenameToObject ObjectMap;  public:     template <typename T>      T *Get () const     {         TypenameToObject::const_iterator iType = ObjectMap.find(&typeid(T));         if (iType == ObjectMap.end())             return NULL;         return reinterpret_cast<T *>(iType->second);     }     template <typename T>      void Set(T *value)      {         ObjectMap[&typeid(T)] = reinterpret_cast<void *>(value);     } };  int main() {     TypeMap Services;     Services.Set<SomeClass>(new SomeClass());     SomeClass *x = Services.Get<SomeClass>(); }  
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1 Answer

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  1. 2026-05-11T02:48:25+00:00Added an answer on May 11, 2026 at 2:48 am

    For this code to compile, the following line:

    typedef map<type_info *, void *, type_info_less> TypenameToObject; 

    should be:

    typedef map<const type_info *, void *, type_info_less> TypenameToObject; 
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