
29 Aug
2015
29 Aug
'15
2:25 a.m.
On Fri, Aug 28, 2015 at 9:47 PM, Agustín K-ballo Bergé <kaballo86@hotmail.com> wrote:
On 8/28/2015 10:35 PM, Nat Goodspeed wrote:
(Is there a simpler way to obtain a variant of the result types of passed functions? I was hoping I could just write: boost::variant< std::result_of<Fns>::type... >.)
So probably this would to:
typedef boost::variant< typename std::result_of<Fns()>::type...> >::type return_t;
Thank you very much! Indeed, this works: typedef boost::variant< typename std::result_of<Fns()>::type... > return_t; So my silly mistake was omitting the parentheses. :-)