Hello Yutaka, The problem is that T is not a template parameter of you function f1. It is a template parameter of the class A. So when the class A is instantiated with some type that is not arithmetic, say void*, it is as if you would directly write a function: enable_if<boost::is_arithmetic<void*>, void>::type which is enable_if<false, void>::type which is an error, since enable_if<false,void> is an empty class. You can write your code as follows to make enable_if work, but it may not be a solution that works: class A { ... template <class T> enable_if< boost::is_arithmetic<T>, void > :: type f1() { ... } }; Also, here you need to explicitly instantiate f1 as A a; a.f1<int>(); Best, Jaakko Järvi On Apr 1, 2005, at 3:27 AM, Yutaka Leon Suematsu wrote:
Dear booster,
I have a question related to enable_if. I would like to enable or disable some member functions from a template class, but enable_if is -not enable- to do it :-)
Here is the code.
template <class T> class A { ... enable_if< boost::is_arithmetic<T>, void > :: type f1() { ... } };
In class A, I want to define function f1 only if the template parameter is arithmetic. It works fine except when the parameter is not arithmetic. I have a compile error "error C2039: 'type' : is not a member of 'boost::enable_if<Cond,T>'" (Working with VC7.1, and boost 3.12)
Thank you in advance for any suggestions and help.
Sincerely yours,
Yutaka Leon Suematsu
_______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users
-- Jaakko Järvi, Assistant Professor -- Texas A&M University, Computer Science -- jarvi@cs.tamu.edu