Why isPrototypeOf() returns false?





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}







6















I have above constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why it's happening? 






function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








javascript







share|improve this question




















  • 1





    Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

    – Kaiido
    51 mins ago











  • Updated my question, sorry about that type

    – Gautam
    49 mins ago











  • try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

    – dandavis
    43 mins ago


















6















I have above constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why it's happening? 






function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








javascript







share|improve this question




















  • 1





    Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

    – Kaiido
    51 mins ago











  • Updated my question, sorry about that type

    – Gautam
    49 mins ago











  • try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

    – dandavis
    43 mins ago














6












6








6


1






I have above constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why it's happening? 






function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








javascript







share|improve this question
















I have above constructors and SubType prototype pointing to an instance of SuperType. When I do x.isPrototypeOf(SubType.prototype) it returns false. I am confused as I have explicitly set x as a prototype for SubType. Can someone tell me why it's happening? 






function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true





function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



console.log(x.isPrototypeOf(SubType)) // returns false

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true





javascript




javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 49 mins ago







Gautam

















asked 1 hour ago









GautamGautam

600413




600413








  • 1





    Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

    – Kaiido
    51 mins ago











  • Updated my question, sorry about that type

    – Gautam
    49 mins ago











  • try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

    – dandavis
    43 mins ago














  • 1





    Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

    – Kaiido
    51 mins ago











  • Updated my question, sorry about that type

    – Gautam
    49 mins ago











  • try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

    – dandavis
    43 mins ago








1




1





Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

– Kaiido
51 mins ago





Not sure to get my head all clear here, but x === SubType.prototype how do you expect it to be its own prototype?

– Kaiido
51 mins ago













Updated my question, sorry about that type

– Gautam
49 mins ago





Updated my question, sorry about that type

– Gautam
49 mins ago













try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
43 mins ago





try console.log(x.isPrototypeOf(new SubType)) for example of how it's used.

– dandavis
43 mins ago












2 Answers
2






active

oldest

votes


















5














SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:






function SuperType(){}

    

function SubType(){}



x = new SuperType();



SubType.prototype = x;

SubType.prototype.constructor = SubType;



const instance = new SubType();

console.log(x.isPrototypeOf(instance)) // returns true

console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








share|improve this answer































    2














    It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:



    function SuperType(foo){ this.foo = foo };
    
function SubType(bar){ this.bar = bar };
    

    
var x = new SubType("bar");
    

    
SuperType.prototype = x;
    
SuperType.prototype.constructor = SubType;


    Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:



    var y = new SuperType("foo");
    
console.log(x.isPrototypeOf(y)) // returns true


    In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.



    console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true





    share|improve this answer








    New contributor




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





















      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%2f55821319%2fwhy-isprototypeof-returns-false%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









      5














      SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:






      function SuperType(){}
      
    
      
function SubType(){}
      

      
x = new SuperType();
      

      
SubType.prototype = x;
      
SubType.prototype.constructor = SubType;
      

      
const instance = new SubType();
      
console.log(x.isPrototypeOf(instance)) // returns true
      
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








      share|improve this answer




























        5














        SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:






        function SuperType(){}
        
    
        
function SubType(){}
        

        
x = new SuperType();
        

        
SubType.prototype = x;
        
SubType.prototype.constructor = SubType;
        

        
const instance = new SubType();
        
console.log(x.isPrototypeOf(instance)) // returns true
        
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








        share|improve this answer


























          5












          5








          5







          SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:






          function SuperType(){}
          
    
          
function SubType(){}
          

          
x = new SuperType();
          

          
SubType.prototype = x;
          
SubType.prototype.constructor = SubType;
          

          
const instance = new SubType();
          
console.log(x.isPrototypeOf(instance)) // returns true
          
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








          share|improve this answer













          SubType is a function. What you probably want to check is if an instance of SubType would inherit from x:






          function SuperType(){}
          
    
          
function SubType(){}
          

          
x = new SuperType();
          

          
SubType.prototype = x;
          
SubType.prototype.constructor = SubType;
          

          
const instance = new SubType();
          
console.log(x.isPrototypeOf(instance)) // returns true
          
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true








          function SuperType(){}
          
    
          
function SubType(){}
          

          
x = new SuperType();
          

          
SubType.prototype = x;
          
SubType.prototype.constructor = SubType;
          

          
const instance = new SubType();
          
console.log(x.isPrototypeOf(instance)) // returns true
          
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true





          function SuperType(){}
          
    
          
function SubType(){}
          

          
x = new SuperType();
          

          
SubType.prototype = x;
          
SubType.prototype.constructor = SubType;
          

          
const instance = new SubType();
          
console.log(x.isPrototypeOf(instance)) // returns true
          
console.log(SuperType.prototype.isPrototypeOf(SubType.prototype)) // returns true






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 43 mins ago









          KaiidoKaiido

          46.4k468109




          46.4k468109

























              2














              It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:



              function SuperType(foo){ this.foo = foo };
              
function SubType(bar){ this.bar = bar };
              

              
var x = new SubType("bar");
              

              
SuperType.prototype = x;
              
SuperType.prototype.constructor = SubType;


              Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:



              var y = new SuperType("foo");
              
console.log(x.isPrototypeOf(y)) // returns true


              In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.



              console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true





              share|improve this answer








              New contributor




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

























                2














                It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:



                function SuperType(foo){ this.foo = foo };
                
function SubType(bar){ this.bar = bar };
                

                
var x = new SubType("bar");
                

                
SuperType.prototype = x;
                
SuperType.prototype.constructor = SubType;


                Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:



                var y = new SuperType("foo");
                
console.log(x.isPrototypeOf(y)) // returns true


                In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.



                console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true





                share|improve this answer








                New contributor




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























                  2












                  2








                  2







                  It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:



                  function SuperType(foo){ this.foo = foo };
                  
function SubType(bar){ this.bar = bar };
                  

                  
var x = new SubType("bar");
                  

                  
SuperType.prototype = x;
                  
SuperType.prototype.constructor = SubType;


                  Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:



                  var y = new SuperType("foo");
                  
console.log(x.isPrototypeOf(y)) // returns true


                  In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.



                  console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true





                  share|improve this answer








                  New contributor




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










                  It helps to add properties to the objects to see what's happening. I fixed a little of your code. You can run this in the console:



                  function SuperType(foo){ this.foo = foo };
                  
function SubType(bar){ this.bar = bar };
                  

                  
var x = new SubType("bar");
                  

                  
SuperType.prototype = x;
                  
SuperType.prototype.constructor = SubType;


                  Now, you asked x.isPrototypeOf(SuperType) and it returns false, because x is not a property of the class SuperType. But when you instantiate a SuperType, x is a property of that new object:



                  var y = new SuperType("foo");
                  
console.log(x.isPrototypeOf(y)) // returns true


                  In your example that is true, SubType.prototype is a prototype of SuperType.prototype and returns true.



                  console.log(SubType.prototype.isPrototypeOf(SuperType.prototype)) // returns true






                  share|improve this answer








                  New contributor




                  David Klinge 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






                  New contributor




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









                  answered 30 mins ago









                  David KlingeDavid Klinge

                  564




                  564




                  New contributor




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





                  New contributor





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






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






























                      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%2f55821319%2fwhy-isprototypeof-returns-false%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

                      What other Star Trek series did the main TNG cast show up in?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
                      Read more

                      Berlina muro

                      Image
                      La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
                      Read more

                      Berlina aerponto

                      Image
                      Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
                      Read more