Re: [Boost-users] How to deduce (possibly void) return type of member function?

Hi Terry, Terry Golubiewski schrieb:

If I have a class...

struct X { int f(double x); void f(int y); };

.... and a template Y that wants to deduce the return-type of some member functions of T...

template<class T> struct Y { typedef typename what_goes_here<T>::type f_double_return_type; typedef typename what_goes here<T>::type f_int_return_type; };

you should use boost::result_of to do such of things. See http://www.boost.org/doc/libs/1_43_0/libs/utility/utility.htm#result_of . Here a description. The class template |result_of| helps determine the type of a call expression. Given an lvalue |f| of type |F| and lvalues |t1|, |t2|, ..., |t/N/| of types |T1|, |T2|, ..., |T/N/|, respectively, the type |result_of::type| defines the result type of the expression |f(t1, t2, ...,t/N/)|. The implementation permits the type |F| to be a function pointer, function reference, member function pointer, or class type. When |F| is a class type with a member type |result_type|, |result_of| is |F::result_type|. Otherwise, |result_of| is |F::result::type| when |/N/ > 0| or |void| when |/N/ = 0|. For additional information about |result_of|, see the C++ Library Technical Report, N1836 http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1836.pdf, or, for motivation and design rationale, the |result_of| proposal http://anubis.dkuug.dk/jtc1/sc22/wg21/docs/papers/2003/n1454.html. Kim

I"ve tried using...

typedef BOOST_TYPEOF_TPL(static_cast(0)->f(double(0))) f_double_type; // this works

.... but ...

typedef BOOST_TYPEOF_TPL(static_cast(0)->f(int(0)) f_int_type; // does not work for "void"

Of course, I wouldn't know which one returns "void" or not ahead of time.

terry

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