Alias for variables in pattern matching












0














I often have something similar to the following (standard type definition for trees):



match tree with
| Branch(v, Branch(vl, tll, tlr), _) = f Branch(vl, tll, tlr)


In other languages there is the ability to do something like:



match tree with
| Branch(v, tl@Branch(_, _, _), _) = f tl


Does OCaml have something similar?










share|improve this question





























    0














    I often have something similar to the following (standard type definition for trees):



    match tree with
    | Branch(v, Branch(vl, tll, tlr), _) = f Branch(vl, tll, tlr)


    In other languages there is the ability to do something like:



    match tree with
    | Branch(v, tl@Branch(_, _, _), _) = f tl


    Does OCaml have something similar?










    share|improve this question



























      0












      0








      0







      I often have something similar to the following (standard type definition for trees):



      match tree with
      | Branch(v, Branch(vl, tll, tlr), _) = f Branch(vl, tll, tlr)


      In other languages there is the ability to do something like:



      match tree with
      | Branch(v, tl@Branch(_, _, _), _) = f tl


      Does OCaml have something similar?










      share|improve this question















      I often have something similar to the following (standard type definition for trees):



      match tree with
      | Branch(v, Branch(vl, tll, tlr), _) = f Branch(vl, tll, tlr)


      In other languages there is the ability to do something like:



      match tree with
      | Branch(v, tl@Branch(_, _, _), _) = f tl


      Does OCaml have something similar?







      ocaml






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 11 at 18:13

























      asked Nov 11 at 18:03









      Cjen1

      8581732




      8581732
























          1 Answer
          1






          active

          oldest

          votes


















          1














          This is done using the as keyword in OCaml:



          match tree with
          | Branch(v, (Branch(_, _, _) as tl), _) = f tl





          share|improve this answer























          • This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
            – Cjen1
            Nov 11 at 18:23






          • 1




            @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
            – sepp2k
            Nov 11 at 18:31











          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',
          autoActivateHeartbeat: false,
          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%2f53251626%2falias-for-variables-in-pattern-matching%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          This is done using the as keyword in OCaml:



          match tree with
          | Branch(v, (Branch(_, _, _) as tl), _) = f tl





          share|improve this answer























          • This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
            – Cjen1
            Nov 11 at 18:23






          • 1




            @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
            – sepp2k
            Nov 11 at 18:31
















          1














          This is done using the as keyword in OCaml:



          match tree with
          | Branch(v, (Branch(_, _, _) as tl), _) = f tl





          share|improve this answer























          • This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
            – Cjen1
            Nov 11 at 18:23






          • 1




            @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
            – sepp2k
            Nov 11 at 18:31














          1












          1








          1






          This is done using the as keyword in OCaml:



          match tree with
          | Branch(v, (Branch(_, _, _) as tl), _) = f tl





          share|improve this answer














          This is done using the as keyword in OCaml:



          match tree with
          | Branch(v, (Branch(_, _, _) as tl), _) = f tl






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 11 at 18:31

























          answered Nov 11 at 18:16









          sepp2k

          292k38593608




          292k38593608












          • This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
            – Cjen1
            Nov 11 at 18:23






          • 1




            @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
            – sepp2k
            Nov 11 at 18:31


















          • This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
            – Cjen1
            Nov 11 at 18:23






          • 1




            @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
            – sepp2k
            Nov 11 at 18:31
















          This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
          – Cjen1
          Nov 11 at 18:23




          This is failing to compile for me, although it works if I surround the Branch(_,_,_) as tl in parenthesis
          – Cjen1
          Nov 11 at 18:23




          1




          1




          @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
          – sepp2k
          Nov 11 at 18:31




          @Cjen1 Right, my mistake. Without the parentheses it take v, Branch(_,_,_) to be the left operand of as.
          – sepp2k
          Nov 11 at 18:31


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Stack Overflow!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53251626%2falias-for-variables-in-pattern-matching%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Full-time equivalent

          Bicuculline

          What is this shape that looks like a rectangle with rounded ends called?