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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                      迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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                      南乌拉尔铁路局

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                      南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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