Is it possible to insert extra operation in fold expression?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
7
down vote

favorite
1












In C++17, fold expression is available, so to print arguments, we could use



template<typename ...Args>
void output_argus(Args&&... args) 

 (cout << ... << args) << EOL;



int main()

 output_argus(1, "test", 5.6f);



having the output
1test5.6



What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?



1
test
5.6


Is that even possible? If yes, how?






c++ templates c++17 variadic-templates fold-expression







share|improve this question



























    up vote
    7
    down vote

    favorite
    1












    In C++17, fold expression is available, so to print arguments, we could use



    template<typename ...Args>
    void output_argus(Args&&... args) 

     (cout << ... << args) << EOL;



    int main()

     output_argus(1, "test", 5.6f);



    having the output
    1test5.6



    What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?



    1
    test
    5.6


    Is that even possible? If yes, how?






    c++ templates c++17 variadic-templates fold-expression







    share|improve this question

























      up vote
      7
      down vote

      favorite
      1









      up vote
      7
      down vote

      favorite
      1






      1





      In C++17, fold expression is available, so to print arguments, we could use



      template<typename ...Args>
      void output_argus(Args&&... args) 

       (cout << ... << args) << EOL;



      int main()

       output_argus(1, "test", 5.6f);



      having the output
      1test5.6



      What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?



      1
      test
      5.6


      Is that even possible? If yes, how?






      c++ templates c++17 variadic-templates fold-expression







      share|improve this question















      In C++17, fold expression is available, so to print arguments, we could use



      template<typename ...Args>
      void output_argus(Args&&... args) 

       (cout << ... << args) << EOL;



      int main()

       output_argus(1, "test", 5.6f);



      having the output
      1test5.6



      What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?



      1
      test
      5.6


      Is that even possible? If yes, how?






      c++ templates c++17 variadic-templates fold-expression




      c++ templates c++17 variadic-templates fold-expression






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 5 hours ago









      max66

      31.7k63459




      31.7k63459










      asked 5 hours ago









      r0ng

      610723




      610723






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          8
          down vote



          accepted











          What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?




          You can use the power of the comma operator



           ((std::cout << args << std::endl), ...);


          or, as suggested by Quentin (thanks) and as you asked, you can simply use n instead of std::endl (to avoid multiple flushing of the stream)



           ((std::cout << args << 'n'), ...);





          share|improve this answer


















          • 2




            std::endl is completely overkill, OP just wants a newline.
            – Quentin
            5 hours ago










          • @Quentin - good point; added to the answer.
            – max66
            5 hours ago

















          up vote
          1
          down vote













          I know that the comma operator is probably the easiest way to do that, but for completeness here's something I came up with, mainly because I wanted to show off my little generalisation of iomanip. The standard library iomanips are functions. There's an << overload that takes a function pointer. I extended that for arbitrary callable objects that take and return streams by reference.



          template <class Stream, class Func>
          auto operator << (Stream& s, Func f) -> 
           std::enable_if_t<std::is_same_v<decltype(f(s)), Stream&>, Stream&>

           return f(s);



          With this little tool in our toolbox, it's easy to write a fold expression that does absolutely anything we want.



          template<typename ...Args>
          void output_args(Args&&... args)

           (std::cout << ... << [&](auto& x)->auto&return x << args << 'n';);



          This technique can be used in scenarios where we need to capture the value of the fold expression, rather than its side effects. The comma operator is less useful in such contexts.






          share|improve this answer




















          • A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
            – Yakk - Adam Nevraumont
            2 hours ago










          • See my solution based off yours.
            – Yakk - Adam Nevraumont
            2 hours ago

















          up vote
          1
          down vote













          This is @n.m.'s solution without the rude global greedy operator<<.



          template<class Os>
          struct chain_stream ;


          now we can do:



           (chain_streamstd::cout << ... << [&](auto& x)x << args << 'n';);


          and it works.






          share|improve this answer




















            Your Answer






            StackExchange.ifUsing("editor", function ()
            StackExchange.using("externalEditor", function ()
            StackExchange.using("snippets", function ()
            StackExchange.snippets.init();
            );
            );
            , "code-snippets");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "1"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53216992%2fis-it-possible-to-insert-extra-operation-in-fold-expression%23new-answer', 'question_page');

            );

            Post as a guest

















            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            8
            down vote



            accepted











            What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?




            You can use the power of the comma operator



             ((std::cout << args << std::endl), ...);


            or, as suggested by Quentin (thanks) and as you asked, you can simply use n instead of std::endl (to avoid multiple flushing of the stream)



             ((std::cout << args << 'n'), ...);





            share|improve this answer


















            • 2




              std::endl is completely overkill, OP just wants a newline.
              – Quentin
              5 hours ago










            • @Quentin - good point; added to the answer.
              – max66
              5 hours ago














            up vote
            8
            down vote



            accepted











            What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?




            You can use the power of the comma operator



             ((std::cout << args << std::endl), ...);


            or, as suggested by Quentin (thanks) and as you asked, you can simply use n instead of std::endl (to avoid multiple flushing of the stream)



             ((std::cout << args << 'n'), ...);





            share|improve this answer


















            • 2




              std::endl is completely overkill, OP just wants a newline.
              – Quentin
              5 hours ago










            • @Quentin - good point; added to the answer.
              – max66
              5 hours ago












            up vote
            8
            down vote



            accepted







            up vote
            8
            down vote



            accepted







            What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?




            You can use the power of the comma operator



             ((std::cout << args << std::endl), ...);


            or, as suggested by Quentin (thanks) and as you asked, you can simply use n instead of std::endl (to avoid multiple flushing of the stream)



             ((std::cout << args << 'n'), ...);





            share|improve this answer















            What if I would like using the fold expression appending an extra character 'n' to each element to get the following results?




            You can use the power of the comma operator



             ((std::cout << args << std::endl), ...);


            or, as suggested by Quentin (thanks) and as you asked, you can simply use n instead of std::endl (to avoid multiple flushing of the stream)



             ((std::cout << args << 'n'), ...);






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 5 hours ago

























            answered 5 hours ago









            max66

            31.7k63459




            31.7k63459







            • 2




              std::endl is completely overkill, OP just wants a newline.
              – Quentin
              5 hours ago










            • @Quentin - good point; added to the answer.
              – max66
              5 hours ago












            • 2




              std::endl is completely overkill, OP just wants a newline.
              – Quentin
              5 hours ago










            • @Quentin - good point; added to the answer.
              – max66
              5 hours ago







            2




            2




            std::endl is completely overkill, OP just wants a newline.
            – Quentin
            5 hours ago




            std::endl is completely overkill, OP just wants a newline.
            – Quentin
            5 hours ago












            @Quentin - good point; added to the answer.
            – max66
            5 hours ago




            @Quentin - good point; added to the answer.
            – max66
            5 hours ago












            up vote
            1
            down vote













            I know that the comma operator is probably the easiest way to do that, but for completeness here's something I came up with, mainly because I wanted to show off my little generalisation of iomanip. The standard library iomanips are functions. There's an << overload that takes a function pointer. I extended that for arbitrary callable objects that take and return streams by reference.



            template <class Stream, class Func>
            auto operator << (Stream& s, Func f) -> 
             std::enable_if_t<std::is_same_v<decltype(f(s)), Stream&>, Stream&>

             return f(s);



            With this little tool in our toolbox, it's easy to write a fold expression that does absolutely anything we want.



            template<typename ...Args>
            void output_args(Args&&... args)

             (std::cout << ... << [&](auto& x)->auto&return x << args << 'n';);



            This technique can be used in scenarios where we need to capture the value of the fold expression, rather than its side effects. The comma operator is less useful in such contexts.






            share|improve this answer




















            • A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
              – Yakk - Adam Nevraumont
              2 hours ago










            • See my solution based off yours.
              – Yakk - Adam Nevraumont
              2 hours ago














            up vote
            1
            down vote













            I know that the comma operator is probably the easiest way to do that, but for completeness here's something I came up with, mainly because I wanted to show off my little generalisation of iomanip. The standard library iomanips are functions. There's an << overload that takes a function pointer. I extended that for arbitrary callable objects that take and return streams by reference.



            template <class Stream, class Func>
            auto operator << (Stream& s, Func f) -> 
             std::enable_if_t<std::is_same_v<decltype(f(s)), Stream&>, Stream&>

             return f(s);



            With this little tool in our toolbox, it's easy to write a fold expression that does absolutely anything we want.



            template<typename ...Args>
            void output_args(Args&&... args)

             (std::cout << ... << [&](auto& x)->auto&return x << args << 'n';);



            This technique can be used in scenarios where we need to capture the value of the fold expression, rather than its side effects. The comma operator is less useful in such contexts.






            share|improve this answer




















            • A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
              – Yakk - Adam Nevraumont
              2 hours ago










            • See my solution based off yours.
              – Yakk - Adam Nevraumont
              2 hours ago












            up vote
            1
            down vote










            up vote
            1
            down vote









            I know that the comma operator is probably the easiest way to do that, but for completeness here's something I came up with, mainly because I wanted to show off my little generalisation of iomanip. The standard library iomanips are functions. There's an << overload that takes a function pointer. I extended that for arbitrary callable objects that take and return streams by reference.



            template <class Stream, class Func>
            auto operator << (Stream& s, Func f) -> 
             std::enable_if_t<std::is_same_v<decltype(f(s)), Stream&>, Stream&>

             return f(s);



            With this little tool in our toolbox, it's easy to write a fold expression that does absolutely anything we want.



            template<typename ...Args>
            void output_args(Args&&... args)

             (std::cout << ... << [&](auto& x)->auto&return x << args << 'n';);



            This technique can be used in scenarios where we need to capture the value of the fold expression, rather than its side effects. The comma operator is less useful in such contexts.






            share|improve this answer












            I know that the comma operator is probably the easiest way to do that, but for completeness here's something I came up with, mainly because I wanted to show off my little generalisation of iomanip. The standard library iomanips are functions. There's an << overload that takes a function pointer. I extended that for arbitrary callable objects that take and return streams by reference.



            template <class Stream, class Func>
            auto operator << (Stream& s, Func f) -> 
             std::enable_if_t<std::is_same_v<decltype(f(s)), Stream&>, Stream&>

             return f(s);



            With this little tool in our toolbox, it's easy to write a fold expression that does absolutely anything we want.



            template<typename ...Args>
            void output_args(Args&&... args)

             (std::cout << ... << [&](auto& x)->auto&return x << args << 'n';);



            This technique can be used in scenarios where we need to capture the value of the fold expression, rather than its side effects. The comma operator is less useful in such contexts.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 4 hours ago









            n.m.

            69.6k882165




            69.6k882165











            • A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
              – Yakk - Adam Nevraumont
              2 hours ago










            • See my solution based off yours.
              – Yakk - Adam Nevraumont
              2 hours ago
















            • A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
              – Yakk - Adam Nevraumont
              2 hours ago










            • See my solution based off yours.
              – Yakk - Adam Nevraumont
              2 hours ago















            A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
            – Yakk - Adam Nevraumont
            2 hours ago




            A free greedy << operator seems rude. Instead, I'd tag one of the types (the function or the stream).
            – Yakk - Adam Nevraumont
            2 hours ago












            See my solution based off yours.
            – Yakk - Adam Nevraumont
            2 hours ago




            See my solution based off yours.
            – Yakk - Adam Nevraumont
            2 hours ago










            up vote
            1
            down vote













            This is @n.m.'s solution without the rude global greedy operator<<.



            template<class Os>
            struct chain_stream ;


            now we can do:



             (chain_streamstd::cout << ... << [&](auto& x)x << args << 'n';);


            and it works.






            share|improve this answer
























              up vote
              1
              down vote













              This is @n.m.'s solution without the rude global greedy operator<<.



              template<class Os>
              struct chain_stream ;


              now we can do:



               (chain_streamstd::cout << ... << [&](auto& x)x << args << 'n';);


              and it works.






              share|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                This is @n.m.'s solution without the rude global greedy operator<<.



                template<class Os>
                struct chain_stream ;


                now we can do:



                 (chain_streamstd::cout << ... << [&](auto& x)x << args << 'n';);


                and it works.






                share|improve this answer












                This is @n.m.'s solution without the rude global greedy operator<<.



                template<class Os>
                struct chain_stream ;


                now we can do:



                 (chain_streamstd::cout << ... << [&](auto& x)x << args << 'n';);


                and it works.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                Yakk - Adam Nevraumont

                176k19179361




                176k19179361



























                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53216992%2fis-it-possible-to-insert-extra-operation-in-fold-expression%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    Long meetings (6-7 hours a day): Being “babysat” by supervisor

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 31 down vote favorite 4 I hold a PhD in mechanical engineering with a 7 year background in self-guided, self-paced research including 2 years as a graduate level course instructor and a subject matter expert for various engineering projects in a university setting in the United States. For the last 10-11 months, I am working in a research industry in France where my task is to debug C++ spaghetti code written over a period of several years by my current supervisor. It is a team of 2: just I and my supervisor. Diurnally, I am posed with a "bug of the day" to get out of this C++ code. My supervisor generally gives me about 15-30 minutes to solve it and then accosts me about whether or not the task is completed and then rinses and repeats this until the end of the day. Off-late, my supervisor has been holding 6-7 hour long "meetin
                    Read more

                    Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite Many fantasy worlds have multiple humanoid species (elves, orcs, humans, etc) living in the same world, in addition to that, they often have the ability to interbreed. And I can't help but question their believability... fantasy-races reproduction interspecies-relations share | improve this question edited Aug 9 at 14:46 Michael Kjörling ♦ 21.1k 10 85 161 asked Aug 9 at 13:13 Chlodio 118 6 closed as unclear what you're asking by MichaelK, Gryphon, John, Vincent, Frostfyre Aug 9 at 15:25 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. 3 Stack Exchange exist to
                    Read more

                    Confectionery

                    Image
                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
                    Read more