How to deduce (possibly void) return type of member function?

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Terry Golubiewski

22 May 2010 22 May '10

2:53 p.m.

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; }; 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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Kim Kuen Tang

22 May 22 May

3:15 p.m.

New subject: 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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Terry Golubiewski

3:31 p.m.

New subject: How to deduce (possibly void) return type of member function?

How? I read the docs you referenced (previously), but I did not see a concrete example. typedef typename boost::result_of< ...what goes here???... (double)>::type f_double_return_type; terry ----- Original Message ----- From: "Kim Kuen Tang" Newsgroups: gmane.comp.lib.boost.user To: Sent: Saturday, May 22, 2010 10:15 AM Subject: Re: 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.

Kim Kuen Tang

5:18 p.m.

New subject: How to deduce (possibly void) return type of member function?

Terry Golubiewski schrieb:

...

How? I read the docs you referenced (previously), but I did not see a concrete example.

i thought it was straightforward, but looking into details it was not the right reference. =-O However, below is a solution which relies on BOOST_TYPEOF. BOOST_TYPEOF will have a problem if the return_type is void. So i replace it with mpl::void_.

...

typedef typename boost::result_of< ...what goes here???... (double)>::type f_double_return_type;

terry

HTH Kim # include # include # include # include # include struct X { int f(int) { return 1; } double f(double) { return 1.0; } void f(char) {} }; template typename boost::mpl::if_< boost::is_same , boost::mpl::void_ , return_type

...

::type deduce_type(return_type (X::*)(Param));

typedef BOOST_TYPEOF(deduce_type<char>(&X::f)) void_; typedef BOOST_TYPEOF(deduce_type<int>(&X::f)) int_; typedef BOOST_TYPEOF(deduce_type<double>(&X::f)) double_; int main() { BOOST_MPL_ASSERT((boost::is_same)); BOOST_MPL_ASSERT((boost::is_same)); BOOST_MPL_ASSERT((boost::is_same)); return 0; }

Terry Golubiewski

10:46 p.m.

New subject: How to deduce (possibly void) return type of member function?

Mr. Kim Kuen Tang, You are awesome! You approach worked! Thank you! However.... I want to detect the function return type of a user-supplied message handler. This non-generic version compiles and runs as expected. #include #include #include <cstddef> struct MessageHandler1 { void f(const char*); }; // MessageHandler1 struct MessageHandler2 { int f(const char*); }; // MessageHandler2 template boost::mpl::identity<R> TypeDeducer(R (T::*)(const char*)); typedef BOOST_TYPEOF(TypeDeducer(&MessageHandler1::f))::type Type1; typedef BOOST_TYPEOF(TypeDeducer(&MessageHandler2::f))::type Type2; #include <iostream> #include <iomanip> #include <typeinfo> using namespace std; int main() { cout << typeid(Type1).name() << endl; cout << typeid(Type2).name() << endl; } // main But this generic version doesn't compile... 1 #include 2 #include 3 4 #include <cstddef> 5 6 struct MessageHandler1 { 7 void f(const char*); 8 }; // MessageHandler1 9 10 struct MessageHandler2 { 11 int f(const char*); 12 }; // MessageHandler2 13 14 template 15 boost::mpl::identity<R> TypeDeducer(R (T::*)(const char*)); 16 17 template<class T> 18 struct GenericDeducer { 19 typedef BOOST_TYPEOF_TPL(TypeDeducer(&T::f))::type type; 20 }; // GenericDeducer 21 22 #include <iostream> 23 #include <iomanip> 24 #include <typeinfo> 25 26 using namespace std; 27 28 int main() { 29 cout << typeid(GenericDeducer<MessageHandler1>::type).name() << endl; 30 cout << typeid(GenericDeducer<MessageHandler2>::type).name() << endl; 31 } // main try3.cpp using native typeof try3.cpp(19) : error C2784: 'boost::mpl::identity<T> TypeDeducer(R (__thiscall T::* )(const char *))' : could not deduce template argument for 'R (__thiscall T::* )(const char *)' from 'T::f' try3.cpp(15) : see declaration of 'TypeDeducer' try3.cpp(20) : see reference to class template instantiation 'GenericDeducer<T>' being compiled

Kim Kuen Tang

23 May 23 May

7:57 a.m.

New subject: How to deduce (possibly void) return type of member function?

Hi Terry, Terry Golubiewski schrieb:

...

But this generic version doesn't compile...

what you want is a "lazy" version of BOOST_TYPEOF. By wrapping your expression into BOOST_TYPEOF_NESTED_TYPEDEF you will delay the calculation until you really need it. The code below compile and work as expected. HTH Kim # include # include # include <cstddef> struct MessageHandler1 { void f(const char*); }; // MessageHandler1 struct MessageHandler2 { int f(const char*); }; // MessageHandler2 template boost::mpl::identity<R> TypeDeducer(R (T::*)(const char*)); template<class X> struct GenericDeducer { BOOST_TYPEOF_NESTED_TYPEDEF_TPL(Lazy,TypeDeducer(&X::f)) typedef typename Lazy::type::type type; }; // GenericDeducer # include <iostream> # include <iomanip> # include <typeinfo> # include # include using namespace std; int main() { cout << typeid(GenericDeducer<MessageHandler1>::type).name() << endl; cout << typeid(GenericDeducer<MessageHandler2>::type).name() << endl; } // main

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