Unable to use object type as method argument in scala shell












0















The following code doesn't' work in scala shell, but it works in IDE, does anyone how can I use object type as method parameter in scala-shell, thanks.



scala> object A {

     | }

defined object A



scala> def f(a:A) :Unit = {

     | }

<console>:63: error: not found: type A

       def f(a:A) :Unit = {





scala scala-repl







share|improve this question























  • Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

    – Yaneeve
    Nov 21 '18 at 13:25
















0















The following code doesn't' work in scala shell, but it works in IDE, does anyone how can I use object type as method parameter in scala-shell, thanks.



scala> object A {

     | }

defined object A



scala> def f(a:A) :Unit = {

     | }

<console>:63: error: not found: type A

       def f(a:A) :Unit = {





scala scala-repl







share|improve this question























  • Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

    – Yaneeve
    Nov 21 '18 at 13:25














0












0








0








The following code doesn't' work in scala shell, but it works in IDE, does anyone how can I use object type as method parameter in scala-shell, thanks.



scala> object A {

     | }

defined object A



scala> def f(a:A) :Unit = {

     | }

<console>:63: error: not found: type A

       def f(a:A) :Unit = {





scala scala-repl







share|improve this question














The following code doesn't' work in scala shell, but it works in IDE, does anyone how can I use object type as method parameter in scala-shell, thanks.



scala> object A {

     | }

defined object A



scala> def f(a:A) :Unit = {

     | }

<console>:63: error: not found: type A

       def f(a:A) :Unit = {





scala scala-repl




scala scala-repl






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 '18 at 12:32









zjffduzjffdu

6,6842579107




6,6842579107













  • Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

    – Yaneeve
    Nov 21 '18 at 13:25



















  • Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

    – Yaneeve
    Nov 21 '18 at 13:25

















Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

– Yaneeve
Nov 21 '18 at 13:25





Just a thought, a function that accepts a unique singleton as its sole argument, is it not just a thunk?

– Yaneeve
Nov 21 '18 at 13:25












2 Answers
2






active

oldest

votes


















1














You can use Object.type like this:



scala> object A {}

defined object A



scala> def f(a: A.type) = println("hello world")

f: (a: A.type)Unit



scala> f(A)

hello world





share|improve this answer































    0














    Just add trait A :



    trait A
    
object A extends A
    
def f(a:A) :Unit = { }





    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',
      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%2f53412123%2funable-to-use-object-type-as-method-argument-in-scala-shell%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      You can use Object.type like this:



      scala> object A {}
      
defined object A
      

      
scala> def f(a: A.type) = println("hello world")
      
f: (a: A.type)Unit
      

      
scala> f(A)
      
hello world





      share|improve this answer




























        1














        You can use Object.type like this:



        scala> object A {}
        
defined object A
        

        
scala> def f(a: A.type) = println("hello world")
        
f: (a: A.type)Unit
        

        
scala> f(A)
        
hello world





        share|improve this answer


























          1












          1








          1







          You can use Object.type like this:



          scala> object A {}
          
defined object A
          

          
scala> def f(a: A.type) = println("hello world")
          
f: (a: A.type)Unit
          

          
scala> f(A)
          
hello world





          share|improve this answer













          You can use Object.type like this:



          scala> object A {}
          
defined object A
          

          
scala> def f(a: A.type) = println("hello world")
          
f: (a: A.type)Unit
          

          
scala> f(A)
          
hello world






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 21 '18 at 12:41









          Mahmoud HanafyMahmoud Hanafy

          97011528




          97011528

























              0














              Just add trait A :



              trait A
              
object A extends A
              
def f(a:A) :Unit = { }





              share|improve this answer




























                0














                Just add trait A :



                trait A
                
object A extends A
                
def f(a:A) :Unit = { }





                share|improve this answer


























                  0












                  0








                  0







                  Just add trait A :



                  trait A
                  
object A extends A
                  
def f(a:A) :Unit = { }





                  share|improve this answer













                  Just add trait A :



                  trait A
                  
object A extends A
                  
def f(a:A) :Unit = { }






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 21 '18 at 12:38









                  Vladimir BerezkinVladimir Berezkin

                  1,65842030




                  1,65842030






























                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53412123%2funable-to-use-object-type-as-method-argument-in-scala-shell%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more