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Function-Tools: switch tail-call to variadic template

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..otherwise unchanged.

NOTE: we need two variants, since lambdas are always const functions,
      while a member pointer to (non)const function would not be captured
      by that overload and thus recurse into the main case and fail there
      with "has no operator()"
This commit is contained in:
Fischlurch 2017-03-19 00:27:59 +01:00
parent 0b7559ce9a
commit dfea57fd02
1 changed files with 11 additions and 20 deletions
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31
research/try.cpp
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@ -64,34 +64,22 @@ namespace lib {
namespace meta { namespace meta {
template< typename RET, class CLASS template< typename RET, class CLASS
, typename A1 , typename...ARGS
> >
struct _Fun<RET (CLASS::*) (A1) > struct _Fun<RET (CLASS::*) (ARGS...) const>
{ {
typedef RET Ret; typedef RET Ret;
typedef Types<CLASS* const, A1> Args; typedef Types<CLASS* const, ARGS...> Args;
typedef RET Sig(CLASS* const, A1); typedef RET Sig(CLASS* const, ARGS...);
}; };
template< typename RET, class CLASS template< typename RET, class CLASS
, typename A1 , typename...ARGS
> >
struct _Fun<RET (CLASS::*) (A1) const> struct _Fun<RET (CLASS::*) (ARGS...) >
{ {
typedef RET Ret; typedef RET Ret;
typedef Types<CLASS* const, A1> Args; typedef Types<CLASS* const, ARGS...> Args;
typedef RET Sig(CLASS* const, A1); typedef RET Sig(CLASS* const, ARGS...);
};
template< typename RET, class CLASS
, typename A1
, typename A2
>
struct _Fun<RET (CLASS::*) (A1,A2) const>
{
typedef RET Ret;
typedef Types<CLASS* const, A1,A2> Args;
typedef RET Sig(CLASS* const, A1,A2);
}; };
}}//namespace lib::meta }}//namespace lib::meta
@ -170,12 +158,15 @@ main (int, char**)
auto lambda = [&](uint ii) { return funk.fun(ii); }; auto lambda = [&](uint ii) { return funk.fun(ii); };
Fun f7{lambda};
showType (funny); showType (funny);
showType (&funny); showType (&funny);
showType (Funky::notfunny); showType (Funky::notfunny);
showType (memfunP); showType (memfunP);
showType (lambda); showType (lambda);
showType (f7);
cout << "\n\n-------\n"; cout << "\n\n-------\n";