Why isPrototypeOf() returns false?





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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
    2 hours ago











  • Updated my question, sorry about that type

    – Gautam
    2 hours ago











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

    – dandavis
    2 hours 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
    2 hours ago











  • Updated my question, sorry about that type

    – Gautam
    2 hours ago











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

    – dandavis
    2 hours 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 2 hours ago







Gautam

















asked 2 hours 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
    2 hours ago











  • Updated my question, sorry about that type

    – Gautam
    2 hours ago











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

    – dandavis
    2 hours 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
    2 hours ago











  • Updated my question, sorry about that type

    – Gautam
    2 hours ago











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

    – dandavis
    2 hours 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
2 hours 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
2 hours ago













Updated my question, sorry about that type

– Gautam
2 hours ago





Updated my question, sorry about that type

– Gautam
2 hours ago













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

– dandavis
2 hours ago





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

– dandavis
2 hours 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








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    David Klinge is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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 2 hours 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 1 hour 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.






























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