It seems to me that eval_if doesn't use lazy evaluation either. The error message quite obviously refers to an attempt of the compiler to evaluate the expression on the 'false' branch, whereas the condition (is_arithmetic<int>) is true to my best knowledge. Thx, Agoston --------------------------------------------------------------------- #include <iostream> #include #include #include #include <vector> using namespace std; using namespace boost; using namespace mpl; template<typename T> struct value_type { typedef typename eval_if< is_arithmetic<T>, identity<T>, identity<typename T::value_type> >::type // ERROR type; }; int _tmain(int argc, _TCHAR* argv[]) { cout << typeid(value_type<int>::type).name() << endl; // ERROR cout << typeid(value_type<double>::type).name() << endl; cout << typeid(value_type< vector<int> >::type).name() << endl; cout << typeid(value_type< vector<double> >::type).name() << endl; return 0; } --------------------------------------------------------------------- Error: error C2825: 'T::value_type': cannot form a qualified name c:\guszti\prog\vcpp\Boost\xEvalIf\xEvalIf.cpp(27) : see reference to class template instantiation 'value_type<T>' being compiled with [ T=int ] error C2143: syntax error : missing ',' before 'value_type' error C2275: 'value_type' : illegal use of this type as an expression "Robert Ramey" az alábbiakat írta a következõ hírüzenetben: cqaid5$sp1$1@sea.gmane.org...

Look in the mpl documentation under eval_if

Robert Ramey

Agoston Bejo wrote:

...

Hi! I'm looking for some "lazy" version of boost::mpl::if_. I'm not very much into MPL, so I

don't really know either if such a function exists in boost or whether it can be easily

constructed. Platform: VC++7.1

Thx in advance, Agoston

This example shows you how it may cause problems that both "branches" of if_ get evaluated:

//------------------------------------------------------------ #include <iostream> #include <vector> #include #include

using namespace std; using namespace boost; using namespace mpl;

template<typename T> struct t_type { typedef typename if_< is_arithmetic<T>, T, typename T::value_type>::type // ERROR! type; };

int _tmain(int argc, _TCHAR* argv[]) { cout << typeid(t_type<int>::type).name() << endl; // 'int' cout << typeid(t_type::type).name() << endl; // 'double' return 0; }

//------------------------------------------------------------ Error message:

error C2825: 'T::value_type': cannot form a qualified name see reference to class template instantiation 't_type<T>' being compiled with [ T=int ]

"Agoston Bejo" wrote

template<typename T> struct value_type { typedef typename eval_if< is_arithmetic<T>, identity<T>, identity<typename T::value_type> >::type

// ERROR

type; };

In your example T::value_type is unconditionally evaluated, and eval_if will not help you in this particular case. You have to ensure that T::value_type is never mentioned in a template that can be possibly instantiated with a type that doesn't have this typedef (eval_if does this for ::type). You can either switch to using ::type instead of ::value_type, and then eval_if will probably help you, or just try something more basic, for example: template struct value_type_impl; template<typename T> struct value_type_impl : identity<T> {}; template<typename T> struct value_type_impl : identity<typename T::value_type> {}; template<typename T> struct value_type : value_type_impl {}; HTH, Arkadiy