What is the need of template lambda introduced in C++20 when C++14 already has generic lambda?












36














c++14 introduced generic lambdas that made it possible to write following:



auto func = (auto a, auto b){
return a + b;
};
auto Foo = func(2, 5);
auto Bar = func("hello", "world");


It is very clear that this generic lambda func works just like a templated function func would work.



Why did the C++ committee decide to add template syntax for generic lamda?










share|improve this question




















  • 4




    What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
    – Some programmer dude
    18 hours ago












  • I've been told this was an interesting use case.
    – Max Langhof
    18 hours ago
















36














c++14 introduced generic lambdas that made it possible to write following:



auto func = (auto a, auto b){
return a + b;
};
auto Foo = func(2, 5);
auto Bar = func("hello", "world");


It is very clear that this generic lambda func works just like a templated function func would work.



Why did the C++ committee decide to add template syntax for generic lamda?










share|improve this question




















  • 4




    What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
    – Some programmer dude
    18 hours ago












  • I've been told this was an interesting use case.
    – Max Langhof
    18 hours ago














36












36








36


6





c++14 introduced generic lambdas that made it possible to write following:



auto func = (auto a, auto b){
return a + b;
};
auto Foo = func(2, 5);
auto Bar = func("hello", "world");


It is very clear that this generic lambda func works just like a templated function func would work.



Why did the C++ committee decide to add template syntax for generic lamda?










share|improve this question















c++14 introduced generic lambdas that made it possible to write following:



auto func = (auto a, auto b){
return a + b;
};
auto Foo = func(2, 5);
auto Bar = func("hello", "world");


It is very clear that this generic lambda func works just like a templated function func would work.



Why did the C++ committee decide to add template syntax for generic lamda?







c++ c++14 c++20






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 14 hours ago









JavaA

1711




1711










asked 18 hours ago









coder3101coder3101

1,219815




1,219815








  • 4




    What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
    – Some programmer dude
    18 hours ago












  • I've been told this was an interesting use case.
    – Max Langhof
    18 hours ago














  • 4




    What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
    – Some programmer dude
    18 hours ago












  • I've been told this was an interesting use case.
    – Max Langhof
    18 hours ago








4




4




What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
– Some programmer dude
18 hours ago






What if you need to use a different template type than for the arguments or return type? And what if it's needed inside the body?
– Some programmer dude
18 hours ago














I've been told this was an interesting use case.
– Max Langhof
18 hours ago




I've been told this was an interesting use case.
– Max Langhof
18 hours ago












4 Answers
4






active

oldest

votes


















54














C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this:



template <class T, class U>
auto operator()(T t, U u) const;


But not like this:



template <class T>
auto operator()(T t1, T t2) const; // Same type please


Nor like this:



template <class T, std::size_t N>
auto operator()(std::array<T, N> const &) const; // Only `std::array` please


Nor like this (although this gets a bit tricky to actually use):



template <class T>
auto operator()() const; // No deduction


C++14 lambdas are fine, but C++20 allows us to implement these cases without hassle.






share|improve this answer





























    18














    Since you can use templated lambdas in C++20, you can restrict your types in an easier way than SFINAE expression:



    auto lambda = <typename T>(std::vector<T> t){};


    This lambda will work only with vector types.






    share|improve this answer



















    • 7




      How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
      – StoryTeller
      18 hours ago










    • It is more an information about what C++20 add to the lambda expression than an answer to the question
      – Antoine Morrier
      18 hours ago



















    10















    The new "familiar template syntax" for lambdas introduced in C++20
    makes constructs such as for_types and for_range viable and way more
    readable compared to C++17 alternatives.




    (source: compile-time iteration with C++20 lambdas)



    Another interesting thing that can be done on both C++14 and C++17 generic lambdas is directly calling operator() by explicitly passing a template parameter:



    C++14:



       auto l = (auto){ };
    l.template operator()<int>(0);


    C++20:



      auto l = <typename T>(){ };
    l.template operator()<int>();


    The C++14 example above is quite useless: there's no way of referring to the type provided to operator() in the body of the lambda without giving the argument a name and using decltype. Additionally, we're forced to pass an argument even though we might not need it.



    The C++20 example shows how T is easily accessible in the body of the lambda and that a nullary lambda can now be arbitrarily templated. This is going to be very useful for the implementation of the aforementioned compile-time constructs






    share|improve this answer










    New contributor




    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


























      7














      The proposal that was accepted into C++20 has a lengthy motivation section, with examples. The premise of it is this:




      There are a few key reasons why the current syntax for defining
      generic lambdas is deemed insufficient by the author. The gist of it is
      that some things that can be done easily with normal function templates
      require significant hoop jumping to be done with generic lambdas, or
      can’t be done at all.The author thinks that lambdas are valuable
      enough that C++ should support them just as well as normal function
      templates.




      Following that are quite a few examples.






      share|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        54














        C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this:



        template <class T, class U>
        auto operator()(T t, U u) const;


        But not like this:



        template <class T>
        auto operator()(T t1, T t2) const; // Same type please


        Nor like this:



        template <class T, std::size_t N>
        auto operator()(std::array<T, N> const &) const; // Only `std::array` please


        Nor like this (although this gets a bit tricky to actually use):



        template <class T>
        auto operator()() const; // No deduction


        C++14 lambdas are fine, but C++20 allows us to implement these cases without hassle.






        share|improve this answer


























          54














          C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this:



          template <class T, class U>
          auto operator()(T t, U u) const;


          But not like this:



          template <class T>
          auto operator()(T t1, T t2) const; // Same type please


          Nor like this:



          template <class T, std::size_t N>
          auto operator()(std::array<T, N> const &) const; // Only `std::array` please


          Nor like this (although this gets a bit tricky to actually use):



          template <class T>
          auto operator()() const; // No deduction


          C++14 lambdas are fine, but C++20 allows us to implement these cases without hassle.






          share|improve this answer
























            54












            54








            54






            C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this:



            template <class T, class U>
            auto operator()(T t, U u) const;


            But not like this:



            template <class T>
            auto operator()(T t1, T t2) const; // Same type please


            Nor like this:



            template <class T, std::size_t N>
            auto operator()(std::array<T, N> const &) const; // Only `std::array` please


            Nor like this (although this gets a bit tricky to actually use):



            template <class T>
            auto operator()() const; // No deduction


            C++14 lambdas are fine, but C++20 allows us to implement these cases without hassle.






            share|improve this answer












            C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this:



            template <class T, class U>
            auto operator()(T t, U u) const;


            But not like this:



            template <class T>
            auto operator()(T t1, T t2) const; // Same type please


            Nor like this:



            template <class T, std::size_t N>
            auto operator()(std::array<T, N> const &) const; // Only `std::array` please


            Nor like this (although this gets a bit tricky to actually use):



            template <class T>
            auto operator()() const; // No deduction


            C++14 lambdas are fine, but C++20 allows us to implement these cases without hassle.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 18 hours ago









            QuentinQuentin

            45k586141




            45k586141

























                18














                Since you can use templated lambdas in C++20, you can restrict your types in an easier way than SFINAE expression:



                auto lambda = <typename T>(std::vector<T> t){};


                This lambda will work only with vector types.






                share|improve this answer



















                • 7




                  How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                  – StoryTeller
                  18 hours ago










                • It is more an information about what C++20 add to the lambda expression than an answer to the question
                  – Antoine Morrier
                  18 hours ago
















                18














                Since you can use templated lambdas in C++20, you can restrict your types in an easier way than SFINAE expression:



                auto lambda = <typename T>(std::vector<T> t){};


                This lambda will work only with vector types.






                share|improve this answer



















                • 7




                  How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                  – StoryTeller
                  18 hours ago










                • It is more an information about what C++20 add to the lambda expression than an answer to the question
                  – Antoine Morrier
                  18 hours ago














                18












                18








                18






                Since you can use templated lambdas in C++20, you can restrict your types in an easier way than SFINAE expression:



                auto lambda = <typename T>(std::vector<T> t){};


                This lambda will work only with vector types.






                share|improve this answer














                Since you can use templated lambdas in C++20, you can restrict your types in an easier way than SFINAE expression:



                auto lambda = <typename T>(std::vector<T> t){};


                This lambda will work only with vector types.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 14 hours ago









                anatolyg

                16.8k44590




                16.8k44590










                answered 18 hours ago









                Antoine MorrierAntoine Morrier

                1,734518




                1,734518








                • 7




                  How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                  – StoryTeller
                  18 hours ago










                • It is more an information about what C++20 add to the lambda expression than an answer to the question
                  – Antoine Morrier
                  18 hours ago














                • 7




                  How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                  – StoryTeller
                  18 hours ago










                • It is more an information about what C++20 add to the lambda expression than an answer to the question
                  – Antoine Morrier
                  18 hours ago








                7




                7




                How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                – StoryTeller
                18 hours ago




                How is consteval related to the new syntax? It's cool and all, but I don't get the relevance.
                – StoryTeller
                18 hours ago












                It is more an information about what C++20 add to the lambda expression than an answer to the question
                – Antoine Morrier
                18 hours ago




                It is more an information about what C++20 add to the lambda expression than an answer to the question
                – Antoine Morrier
                18 hours ago











                10















                The new "familiar template syntax" for lambdas introduced in C++20
                makes constructs such as for_types and for_range viable and way more
                readable compared to C++17 alternatives.




                (source: compile-time iteration with C++20 lambdas)



                Another interesting thing that can be done on both C++14 and C++17 generic lambdas is directly calling operator() by explicitly passing a template parameter:



                C++14:



                   auto l = (auto){ };
                l.template operator()<int>(0);


                C++20:



                  auto l = <typename T>(){ };
                l.template operator()<int>();


                The C++14 example above is quite useless: there's no way of referring to the type provided to operator() in the body of the lambda without giving the argument a name and using decltype. Additionally, we're forced to pass an argument even though we might not need it.



                The C++20 example shows how T is easily accessible in the body of the lambda and that a nullary lambda can now be arbitrarily templated. This is going to be very useful for the implementation of the aforementioned compile-time constructs






                share|improve this answer










                New contributor




                Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.























                  10















                  The new "familiar template syntax" for lambdas introduced in C++20
                  makes constructs such as for_types and for_range viable and way more
                  readable compared to C++17 alternatives.




                  (source: compile-time iteration with C++20 lambdas)



                  Another interesting thing that can be done on both C++14 and C++17 generic lambdas is directly calling operator() by explicitly passing a template parameter:



                  C++14:



                     auto l = (auto){ };
                  l.template operator()<int>(0);


                  C++20:



                    auto l = <typename T>(){ };
                  l.template operator()<int>();


                  The C++14 example above is quite useless: there's no way of referring to the type provided to operator() in the body of the lambda without giving the argument a name and using decltype. Additionally, we're forced to pass an argument even though we might not need it.



                  The C++20 example shows how T is easily accessible in the body of the lambda and that a nullary lambda can now be arbitrarily templated. This is going to be very useful for the implementation of the aforementioned compile-time constructs






                  share|improve this answer










                  New contributor




                  Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                    10












                    10








                    10







                    The new "familiar template syntax" for lambdas introduced in C++20
                    makes constructs such as for_types and for_range viable and way more
                    readable compared to C++17 alternatives.




                    (source: compile-time iteration with C++20 lambdas)



                    Another interesting thing that can be done on both C++14 and C++17 generic lambdas is directly calling operator() by explicitly passing a template parameter:



                    C++14:



                       auto l = (auto){ };
                    l.template operator()<int>(0);


                    C++20:



                      auto l = <typename T>(){ };
                    l.template operator()<int>();


                    The C++14 example above is quite useless: there's no way of referring to the type provided to operator() in the body of the lambda without giving the argument a name and using decltype. Additionally, we're forced to pass an argument even though we might not need it.



                    The C++20 example shows how T is easily accessible in the body of the lambda and that a nullary lambda can now be arbitrarily templated. This is going to be very useful for the implementation of the aforementioned compile-time constructs






                    share|improve this answer










                    New contributor




                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.










                    The new "familiar template syntax" for lambdas introduced in C++20
                    makes constructs such as for_types and for_range viable and way more
                    readable compared to C++17 alternatives.




                    (source: compile-time iteration with C++20 lambdas)



                    Another interesting thing that can be done on both C++14 and C++17 generic lambdas is directly calling operator() by explicitly passing a template parameter:



                    C++14:



                       auto l = (auto){ };
                    l.template operator()<int>(0);


                    C++20:



                      auto l = <typename T>(){ };
                    l.template operator()<int>();


                    The C++14 example above is quite useless: there's no way of referring to the type provided to operator() in the body of the lambda without giving the argument a name and using decltype. Additionally, we're forced to pass an argument even though we might not need it.



                    The C++20 example shows how T is easily accessible in the body of the lambda and that a nullary lambda can now be arbitrarily templated. This is going to be very useful for the implementation of the aforementioned compile-time constructs







                    share|improve this answer










                    New contributor




                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|improve this answer



                    share|improve this answer








                    edited 17 hours ago









                    Toby Speight

                    16.4k133965




                    16.4k133965






                    New contributor




                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 18 hours ago









                    Hamza.SHamza.S

                    383210




                    383210




                    New contributor




                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Hamza.S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.























                        7














                        The proposal that was accepted into C++20 has a lengthy motivation section, with examples. The premise of it is this:




                        There are a few key reasons why the current syntax for defining
                        generic lambdas is deemed insufficient by the author. The gist of it is
                        that some things that can be done easily with normal function templates
                        require significant hoop jumping to be done with generic lambdas, or
                        can’t be done at all.The author thinks that lambdas are valuable
                        enough that C++ should support them just as well as normal function
                        templates.




                        Following that are quite a few examples.






                        share|improve this answer


























                          7














                          The proposal that was accepted into C++20 has a lengthy motivation section, with examples. The premise of it is this:




                          There are a few key reasons why the current syntax for defining
                          generic lambdas is deemed insufficient by the author. The gist of it is
                          that some things that can be done easily with normal function templates
                          require significant hoop jumping to be done with generic lambdas, or
                          can’t be done at all.The author thinks that lambdas are valuable
                          enough that C++ should support them just as well as normal function
                          templates.




                          Following that are quite a few examples.






                          share|improve this answer
























                            7












                            7








                            7






                            The proposal that was accepted into C++20 has a lengthy motivation section, with examples. The premise of it is this:




                            There are a few key reasons why the current syntax for defining
                            generic lambdas is deemed insufficient by the author. The gist of it is
                            that some things that can be done easily with normal function templates
                            require significant hoop jumping to be done with generic lambdas, or
                            can’t be done at all.The author thinks that lambdas are valuable
                            enough that C++ should support them just as well as normal function
                            templates.




                            Following that are quite a few examples.






                            share|improve this answer












                            The proposal that was accepted into C++20 has a lengthy motivation section, with examples. The premise of it is this:




                            There are a few key reasons why the current syntax for defining
                            generic lambdas is deemed insufficient by the author. The gist of it is
                            that some things that can be done easily with normal function templates
                            require significant hoop jumping to be done with generic lambdas, or
                            can’t be done at all.The author thinks that lambdas are valuable
                            enough that C++ should support them just as well as normal function
                            templates.




                            Following that are quite a few examples.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 18 hours ago









                            StoryTellerStoryTeller

                            94.5k12191254




                            94.5k12191254






























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