Simplest way to determine return type of function











up vote
40
down vote

favorite
10












Given a very simple, but lengthy function, such as:



int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???


What is the most simple and concise way to determine the function's return type (ReturnTypeOfFoo, in this example: int) at compile time without repeating the function's parameter types (by name only, since it is known that the function does not have any additional overloads)?






c++ function c++17 return-type compile-time







share|improve this question
























  • Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
    – Jay
    Dec 7 at 16:33








  • 1




    Maybe decltype(foo)::result_type?
    – Thomas Lang
    Dec 7 at 16:33










  • maybe en.cppreference.com/w/cpp/types/result_of?
    – bracco23
    Dec 7 at 16:35






  • 1




    @ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
    – user463035818
    Dec 7 at 16:39










  • @user463035818 You may be right here, I was referring to the result_type member of a std::function type.
    – Thomas Lang
    Dec 7 at 16:40















up vote
40
down vote

favorite
10












Given a very simple, but lengthy function, such as:



int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???


What is the most simple and concise way to determine the function's return type (ReturnTypeOfFoo, in this example: int) at compile time without repeating the function's parameter types (by name only, since it is known that the function does not have any additional overloads)?






c++ function c++17 return-type compile-time







share|improve this question
























  • Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
    – Jay
    Dec 7 at 16:33








  • 1




    Maybe decltype(foo)::result_type?
    – Thomas Lang
    Dec 7 at 16:33










  • maybe en.cppreference.com/w/cpp/types/result_of?
    – bracco23
    Dec 7 at 16:35






  • 1




    @ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
    – user463035818
    Dec 7 at 16:39










  • @user463035818 You may be right here, I was referring to the result_type member of a std::function type.
    – Thomas Lang
    Dec 7 at 16:40













up vote
40
down vote

favorite
10









up vote
40
down vote

favorite
10






10





Given a very simple, but lengthy function, such as:



int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???


What is the most simple and concise way to determine the function's return type (ReturnTypeOfFoo, in this example: int) at compile time without repeating the function's parameter types (by name only, since it is known that the function does not have any additional overloads)?






c++ function c++17 return-type compile-time







share|improve this question















Given a very simple, but lengthy function, such as:



int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???


What is the most simple and concise way to determine the function's return type (ReturnTypeOfFoo, in this example: int) at compile time without repeating the function's parameter types (by name only, since it is known that the function does not have any additional overloads)?






c++ function c++17 return-type compile-time




c++ function c++17 return-type compile-time






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 7 at 17:44









Konrad Rudolph

393k1017731022




393k1017731022










asked Dec 7 at 16:29









Cybran

965717




965717












  • Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
    – Jay
    Dec 7 at 16:33








  • 1




    Maybe decltype(foo)::result_type?
    – Thomas Lang
    Dec 7 at 16:33










  • maybe en.cppreference.com/w/cpp/types/result_of?
    – bracco23
    Dec 7 at 16:35






  • 1




    @ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
    – user463035818
    Dec 7 at 16:39










  • @user463035818 You may be right here, I was referring to the result_type member of a std::function type.
    – Thomas Lang
    Dec 7 at 16:40


















  • Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
    – Jay
    Dec 7 at 16:33








  • 1




    Maybe decltype(foo)::result_type?
    – Thomas Lang
    Dec 7 at 16:33










  • maybe en.cppreference.com/w/cpp/types/result_of?
    – bracco23
    Dec 7 at 16:35






  • 1




    @ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
    – user463035818
    Dec 7 at 16:39










  • @user463035818 You may be right here, I was referring to the result_type member of a std::function type.
    – Thomas Lang
    Dec 7 at 16:40
















Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
– Jay
Dec 7 at 16:33






Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.
– Jay
Dec 7 at 16:33






1




1




Maybe decltype(foo)::result_type?
– Thomas Lang
Dec 7 at 16:33




Maybe decltype(foo)::result_type?
– Thomas Lang
Dec 7 at 16:33












maybe en.cppreference.com/w/cpp/types/result_of?
– bracco23
Dec 7 at 16:35




maybe en.cppreference.com/w/cpp/types/result_of?
– bracco23
Dec 7 at 16:35




1




1




@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
– user463035818
Dec 7 at 16:39




@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?
– user463035818
Dec 7 at 16:39












@user463035818 You may be right here, I was referring to the result_type member of a std::function type.
– Thomas Lang
Dec 7 at 16:40




@user463035818 You may be right here, I was referring to the result_type member of a std::function type.
– Thomas Lang
Dec 7 at 16:40












3 Answers
3






active

oldest

votes

















up vote
48
down vote



accepted










You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:



using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;



We can make this a little more generic like



template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;


which then lets you use it like



int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;





share|improve this answer























  • Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
    – Cybran
    Dec 7 at 19:08


















up vote
19
down vote













Most simple and concise is probably:



template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));


Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.



But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.






share|improve this answer




























    up vote
    13
    down vote













    I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:



    template <typename R, typename ... Args>
    
R getRetType (R(*)(Args...));


    and using decltype()?



    using ReturnTypeOfFoo = decltype( getRetType(&foo) );


    Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.






    share|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      48
      down vote



      accepted










      You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:



      using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;



      We can make this a little more generic like



      template<typename Callable>
      
using return_type_of_t = 
      
    typename decltype(std::function{std::declval<Callable>()})::result_type;


      which then lets you use it like



      int foo(int a, int b, int c, int d) {
      
    return 1;
      
}
      

      
auto bar = (){ return 1; };
      

      
struct baz_ 
      
{ 
      
    double operator()(){ return 0; } 
      
} baz;
      

      
using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;
      
using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;
      
using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;





      share|improve this answer























      • Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
        – Cybran
        Dec 7 at 19:08















      up vote
      48
      down vote



      accepted










      You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:



      using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;



      We can make this a little more generic like



      template<typename Callable>
      
using return_type_of_t = 
      
    typename decltype(std::function{std::declval<Callable>()})::result_type;


      which then lets you use it like



      int foo(int a, int b, int c, int d) {
      
    return 1;
      
}
      

      
auto bar = (){ return 1; };
      

      
struct baz_ 
      
{ 
      
    double operator()(){ return 0; } 
      
} baz;
      

      
using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;
      
using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;
      
using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;





      share|improve this answer























      • Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
        – Cybran
        Dec 7 at 19:08













      up vote
      48
      down vote



      accepted







      up vote
      48
      down vote



      accepted






      You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:



      using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;



      We can make this a little more generic like



      template<typename Callable>
      
using return_type_of_t = 
      
    typename decltype(std::function{std::declval<Callable>()})::result_type;


      which then lets you use it like



      int foo(int a, int b, int c, int d) {
      
    return 1;
      
}
      

      
auto bar = (){ return 1; };
      

      
struct baz_ 
      
{ 
      
    double operator()(){ return 0; } 
      
} baz;
      

      
using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;
      
using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;
      
using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;





      share|improve this answer














      You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:



      using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;



      We can make this a little more generic like



      template<typename Callable>
      
using return_type_of_t = 
      
    typename decltype(std::function{std::declval<Callable>()})::result_type;


      which then lets you use it like



      int foo(int a, int b, int c, int d) {
      
    return 1;
      
}
      

      
auto bar = (){ return 1; };
      

      
struct baz_ 
      
{ 
      
    double operator()(){ return 0; } 
      
} baz;
      

      
using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;
      
using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;
      
using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Dec 8 at 8:50









      coldspeed

      114k18105182




      114k18105182










      answered Dec 7 at 16:44









      NathanOliver

      85.3k15118177




      85.3k15118177












      • Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
        – Cybran
        Dec 7 at 19:08


















      • Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
        – Cybran
        Dec 7 at 19:08
















      Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
      – Cybran
      Dec 7 at 19:08




      Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.
      – Cybran
      Dec 7 at 19:08












      up vote
      19
      down vote













      Most simple and concise is probably:



      template <typename R, typename... Args>
      
R return_type_of(R(*)(Args...));
      

      
using ReturnTypeOfFoo = decltype(return_type_of(foo));


      Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.



      But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.






      share|improve this answer

























        up vote
        19
        down vote













        Most simple and concise is probably:



        template <typename R, typename... Args>
        
R return_type_of(R(*)(Args...));
        

        
using ReturnTypeOfFoo = decltype(return_type_of(foo));


        Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.



        But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.






        share|improve this answer























          up vote
          19
          down vote










          up vote
          19
          down vote









          Most simple and concise is probably:



          template <typename R, typename... Args>
          
R return_type_of(R(*)(Args...));
          

          
using ReturnTypeOfFoo = decltype(return_type_of(foo));


          Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.



          But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.






          share|improve this answer












          Most simple and concise is probably:



          template <typename R, typename... Args>
          
R return_type_of(R(*)(Args...));
          

          
using ReturnTypeOfFoo = decltype(return_type_of(foo));


          Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.



          But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 7 at 16:34









          Barry

          175k18299554




          175k18299554






















              up vote
              13
              down vote













              I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:



              template <typename R, typename ... Args>
              
R getRetType (R(*)(Args...));


              and using decltype()?



              using ReturnTypeOfFoo = decltype( getRetType(&foo) );


              Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.






              share|improve this answer



























                up vote
                13
                down vote













                I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:



                template <typename R, typename ... Args>
                
R getRetType (R(*)(Args...));


                and using decltype()?



                using ReturnTypeOfFoo = decltype( getRetType(&foo) );


                Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.






                share|improve this answer

























                  up vote
                  13
                  down vote










                  up vote
                  13
                  down vote









                  I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:



                  template <typename R, typename ... Args>
                  
R getRetType (R(*)(Args...));


                  and using decltype()?



                  using ReturnTypeOfFoo = decltype( getRetType(&foo) );


                  Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.






                  share|improve this answer














                  I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:



                  template <typename R, typename ... Args>
                  
R getRetType (R(*)(Args...));


                  and using decltype()?



                  using ReturnTypeOfFoo = decltype( getRetType(&foo) );


                  Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 7 at 17:26









                  NathanOliver

                  85.3k15118177




                  85.3k15118177










                  answered Dec 7 at 16:34









                  max66

                  33.8k63762




                  33.8k63762






























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                      格利澤436b

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                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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