Is it possible to use function output in declaration of another function in C++

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











up vote
7
down vote

favorite












I want something like this:



std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);


Obviously in this example code pointer to function would be passed



What I want is to have equivalent of this:



void MyFunction_2 (std::tuple<int, bool, double> &params);


Is it possible to do so?






c++ c++11 decltype







share|improve this question



























    up vote
    7
    down vote

    favorite












    I want something like this:



    std::tuple<int, bool, double> MyFunction_1 (void);

    void MyFunction_2 (decltype (MyFunction_1) &params);


    Obviously in this example code pointer to function would be passed



    What I want is to have equivalent of this:



    void MyFunction_2 (std::tuple<int, bool, double> &params);


    Is it possible to do so?






    c++ c++11 decltype







    share|improve this question

























      up vote
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite











      I want something like this:



      std::tuple<int, bool, double> MyFunction_1 (void);

      void MyFunction_2 (decltype (MyFunction_1) &params);


      Obviously in this example code pointer to function would be passed



      What I want is to have equivalent of this:



      void MyFunction_2 (std::tuple<int, bool, double> &params);


      Is it possible to do so?






      c++ c++11 decltype







      share|improve this question















      I want something like this:



      std::tuple<int, bool, double> MyFunction_1 (void);

      void MyFunction_2 (decltype (MyFunction_1) &params);


      Obviously in this example code pointer to function would be passed



      What I want is to have equivalent of this:



      void MyFunction_2 (std::tuple<int, bool, double> &params);


      Is it possible to do so?






      c++ c++11 decltype




      c++ c++11 decltype






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 24 mins ago









      songyuanyao

      86k9163224




      86k9163224










      asked 37 mins ago









      Dmitrii Motorygin

      555




      555






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          9
          down vote













          decltype (MyFunction_1) will give the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



          void MyFunction_2 (decltype (MyFunction_1()) &params);
          // ^^


          1 The expression is evaluated at compile-time, the function won't be called at run-time.






          share|improve this answer


















          • 1




            Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
            – StoryTeller
            24 mins ago






          • 1




            Would "you need to emulate a function call in the decltype()" be a better wording? :P
            – Rerito
            13 mins ago










          • Another approach (maybe more straightforward): std::result_of
            – Ptaq666
            10 mins ago






          • 2




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago

















          up vote
          6
          down vote













          The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



          So decltype(MyFunction_1) is not what you want.



          The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






          share|improve this answer
















          • 1




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago










          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: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f52811378%2fis-it-possible-to-use-function-output-in-declaration-of-another-function-in-c%23new-answer', 'question_page');

          );

          Post as a guest

















          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          9
          down vote













          decltype (MyFunction_1) will give the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



          void MyFunction_2 (decltype (MyFunction_1()) &params);
          // ^^


          1 The expression is evaluated at compile-time, the function won't be called at run-time.






          share|improve this answer


















          • 1




            Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
            – StoryTeller
            24 mins ago






          • 1




            Would "you need to emulate a function call in the decltype()" be a better wording? :P
            – Rerito
            13 mins ago










          • Another approach (maybe more straightforward): std::result_of
            – Ptaq666
            10 mins ago






          • 2




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago














          up vote
          9
          down vote













          decltype (MyFunction_1) will give the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



          void MyFunction_2 (decltype (MyFunction_1()) &params);
          // ^^


          1 The expression is evaluated at compile-time, the function won't be called at run-time.






          share|improve this answer


















          • 1




            Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
            – StoryTeller
            24 mins ago






          • 1




            Would "you need to emulate a function call in the decltype()" be a better wording? :P
            – Rerito
            13 mins ago










          • Another approach (maybe more straightforward): std::result_of
            – Ptaq666
            10 mins ago






          • 2




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago












          up vote
          9
          down vote










          up vote
          9
          down vote









          decltype (MyFunction_1) will give the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



          void MyFunction_2 (decltype (MyFunction_1()) &params);
          // ^^


          1 The expression is evaluated at compile-time, the function won't be called at run-time.






          share|improve this answer














          decltype (MyFunction_1) will give the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



          void MyFunction_2 (decltype (MyFunction_1()) &params);
          // ^^


          1 The expression is evaluated at compile-time, the function won't be called at run-time.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 mins ago

























          answered 35 mins ago









          songyuanyao

          86k9163224




          86k9163224







          • 1




            Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
            – StoryTeller
            24 mins ago






          • 1




            Would "you need to emulate a function call in the decltype()" be a better wording? :P
            – Rerito
            13 mins ago










          • Another approach (maybe more straightforward): std::result_of
            – Ptaq666
            10 mins ago






          • 2




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago












          • 1




            Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
            – StoryTeller
            24 mins ago






          • 1




            Would "you need to emulate a function call in the decltype()" be a better wording? :P
            – Rerito
            13 mins ago










          • Another approach (maybe more straightforward): std::result_of
            – Ptaq666
            10 mins ago






          • 2




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago







          1




          1




          Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
          – StoryTeller
          24 mins ago




          Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
          – StoryTeller
          24 mins ago




          1




          1




          Would "you need to emulate a function call in the decltype()" be a better wording? :P
          – Rerito
          13 mins ago




          Would "you need to emulate a function call in the decltype()" be a better wording? :P
          – Rerito
          13 mins ago












          Another approach (maybe more straightforward): std::result_of
          – Ptaq666
          10 mins ago




          Another approach (maybe more straightforward): std::result_of
          – Ptaq666
          10 mins ago




          2




          2




          Thanks for the help!
          – Dmitrii Motorygin
          3 mins ago




          Thanks for the help!
          – Dmitrii Motorygin
          3 mins ago












          up vote
          6
          down vote













          The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



          So decltype(MyFunction_1) is not what you want.



          The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






          share|improve this answer
















          • 1




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago














          up vote
          6
          down vote













          The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



          So decltype(MyFunction_1) is not what you want.



          The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






          share|improve this answer
















          • 1




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago












          up vote
          6
          down vote










          up vote
          6
          down vote









          The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



          So decltype(MyFunction_1) is not what you want.



          The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






          share|improve this answer












          The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



          So decltype(MyFunction_1) is not what you want.



          The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 17 mins ago









          Bathsheba

          170k26239360




          170k26239360







          • 1




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago












          • 1




            Thanks for the help!
            – Dmitrii Motorygin
            3 mins ago







          1




          1




          Thanks for the help!
          – Dmitrii Motorygin
          3 mins ago




          Thanks for the help!
          – Dmitrii Motorygin
          3 mins ago

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52811378%2fis-it-possible-to-use-function-output-in-declaration-of-another-function-in-c%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What does second last employer means? [closed]

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
          Read more

          List of Gilmore Girls characters

          Image
          The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
          Read more

          One-line joke

          A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
          Read more