Can the return type of the function be obtained from within the function?

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38
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Can the return type of a function be obtained in a simple way within the function?



For example, given:



template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type 
 typename std::remove_reference<decltype(*p)>::type f; // <-- here

 ...



In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?






c++ c++11 trailing-return-type







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    In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
    – jingyu9575
    Sep 6 at 4:42














up vote
38
down vote

favorite
3












Can the return type of a function be obtained in a simple way within the function?



For example, given:



template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type 
 typename std::remove_reference<decltype(*p)>::type f; // <-- here

 ...



In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?






c++ c++11 trailing-return-type







share|improve this question



















  • 1




    In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
    – jingyu9575
    Sep 6 at 4:42












up vote
38
down vote

favorite
3









up vote
38
down vote

favorite
3






3





Can the return type of a function be obtained in a simple way within the function?



For example, given:



template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type 
 typename std::remove_reference<decltype(*p)>::type f; // <-- here

 ...



In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?






c++ c++11 trailing-return-type







share|improve this question















Can the return type of a function be obtained in a simple way within the function?



For example, given:



template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type 
 typename std::remove_reference<decltype(*p)>::type f; // <-- here

 ...



In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?






c++ c++11 trailing-return-type




c++ c++11 trailing-return-type






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Sep 5 at 18:24

























asked Sep 5 at 17:33









BeeOnRope

24k873162




24k873162







  • 1




    In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
    – jingyu9575
    Sep 6 at 4:42












  • 1




    In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
    – jingyu9575
    Sep 6 at 4:42







1




1




In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
– jingyu9575
Sep 6 at 4:42




In C++14, you can remove the trailing return type instead, and use return f; to deduce the return type.
– jingyu9575
Sep 6 at 4:42












1 Answer
1






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Call the function with a decltype.



decltype(foo(p)) f;





share|improve this answer
















  • 6




    life made easy by decltype ;)
    – JeJo
    Sep 5 at 18:18










  • Works well in this case where there are few parameters...
    – Max Langhof
    Sep 6 at 8:01










  • @MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
    – Rakete1111
    Sep 6 at 8:02










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
50
down vote



accepted










Call the function with a decltype.



decltype(foo(p)) f;





share|improve this answer
















  • 6




    life made easy by decltype ;)
    – JeJo
    Sep 5 at 18:18










  • Works well in this case where there are few parameters...
    – Max Langhof
    Sep 6 at 8:01










  • @MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
    – Rakete1111
    Sep 6 at 8:02














up vote
50
down vote



accepted










Call the function with a decltype.



decltype(foo(p)) f;





share|improve this answer
















  • 6




    life made easy by decltype ;)
    – JeJo
    Sep 5 at 18:18










  • Works well in this case where there are few parameters...
    – Max Langhof
    Sep 6 at 8:01










  • @MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
    – Rakete1111
    Sep 6 at 8:02












up vote
50
down vote



accepted







up vote
50
down vote



accepted






Call the function with a decltype.



decltype(foo(p)) f;





share|improve this answer












Call the function with a decltype.



decltype(foo(p)) f;






share|improve this answer












share|improve this answer



share|improve this answer










answered Sep 5 at 17:35









Rakete1111

32.7k975110




32.7k975110







  • 6




    life made easy by decltype ;)
    – JeJo
    Sep 5 at 18:18










  • Works well in this case where there are few parameters...
    – Max Langhof
    Sep 6 at 8:01










  • @MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
    – Rakete1111
    Sep 6 at 8:02












  • 6




    life made easy by decltype ;)
    – JeJo
    Sep 5 at 18:18










  • Works well in this case where there are few parameters...
    – Max Langhof
    Sep 6 at 8:01










  • @MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
    – Rakete1111
    Sep 6 at 8:02







6




6




life made easy by decltype ;)
– JeJo
Sep 5 at 18:18




life made easy by decltype ;)
– JeJo
Sep 5 at 18:18












Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01




Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01












@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02




@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02

















 

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