Re: [Boost-users] Conditional function definition with boost::enable_if

Hi Matthieu, On Wed, Feb 11, 2009 at 09:22:47AM +0100, Matthieu Brucher wrote:

template<class Child> struct MyfStruct { typename boost::enable_if::type f() { // Do some stuff, but I don't know how to call i(), you will have to test :| } };

where has_i<> tests if the child has the method i().

Then, if you find how to call i() from the f() method, you just have to write:

class MyClass: public MyfStruct<MyClass> { void i(); // f() will be callable now };

Thanks for your example. Yet I believe, that the solution you propose does not solve the underlying problem: as the compiler errors indicate, struct bar (or in your example, class MyClass) is incomplete at the time when has_i<Child> [*] is called. Hence, bar/MyClass cannot use CRTP to automatically derive functions from foo/MyfStruct. I will have to use a different approach to inject functions conditionally. Matthias [*] This would factor out the code that checks whether T has i(): template <typename T> struct has_i : boost::mpl::bool_ {}; -- Matthias Vallentin vallentin@icsi.berkeley.edu http://matthias.vallentin.cc