Why is the 'in' operator throwing an error with a string literal instead of logging false?





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21















As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?






let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)








var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);












share|improve this question

























  • I'd assume the temporary wrapper object created for the string is not enumerable ..?

    – Teemu
    Apr 4 at 18:49













  • @Teemu No. There is no temporary wrapper object created at all

    – Bergi
    Apr 4 at 18:51











  • @Bergi Well, that explains a lot.

    – Teemu
    Apr 4 at 18:52


















21















As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?






let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)








var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);












share|improve this question

























  • I'd assume the temporary wrapper object created for the string is not enumerable ..?

    – Teemu
    Apr 4 at 18:49













  • @Teemu No. There is no temporary wrapper object created at all

    – Bergi
    Apr 4 at 18:51











  • @Bergi Well, that explains a lot.

    – Teemu
    Apr 4 at 18:52














21












21








21








As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?






let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)








var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);












share|improve this question
















As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?






let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)








var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);








let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)





let let1 = new String('test');
console.log(let1.length);
console.log('length' in let1)





var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);





var let1 = 'test';
console.log(let1.length);
console.log('length' in let1);






javascript string






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 4 at 21:55









Boann

37.5k1290122




37.5k1290122










asked Apr 4 at 18:44









brkbrk

30.2k32246




30.2k32246













  • I'd assume the temporary wrapper object created for the string is not enumerable ..?

    – Teemu
    Apr 4 at 18:49













  • @Teemu No. There is no temporary wrapper object created at all

    – Bergi
    Apr 4 at 18:51











  • @Bergi Well, that explains a lot.

    – Teemu
    Apr 4 at 18:52



















  • I'd assume the temporary wrapper object created for the string is not enumerable ..?

    – Teemu
    Apr 4 at 18:49













  • @Teemu No. There is no temporary wrapper object created at all

    – Bergi
    Apr 4 at 18:51











  • @Bergi Well, that explains a lot.

    – Teemu
    Apr 4 at 18:52

















I'd assume the temporary wrapper object created for the string is not enumerable ..?

– Teemu
Apr 4 at 18:49







I'd assume the temporary wrapper object created for the string is not enumerable ..?

– Teemu
Apr 4 at 18:49















@Teemu No. There is no temporary wrapper object created at all

– Bergi
Apr 4 at 18:51





@Teemu No. There is no temporary wrapper object created at all

– Bergi
Apr 4 at 18:51













@Bergi Well, that explains a lot.

– Teemu
Apr 4 at 18:52





@Bergi Well, that explains a lot.

– Teemu
Apr 4 at 18:52












5 Answers
5






active

oldest

votes


















39














In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).



See Distinction between string primitives and String objects



Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.




You must specify an object on the right side of the in operator. For
example, you can specify a string created with the String constructor,
but you cannot specify a string literal.







share|improve this answer





















  • 3





    This is the shortest and most concise correct answer shown.

    – Scott Marcus
    Apr 4 at 19:05











  • Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

    – ChrisM
    Apr 5 at 16:12





















10














It throws an error because in is an operator for objects:




prop in object




but when you declare a string as `` (` string(template) literals) or "" '' (",' string literals) you don't create an object.



Check



typeof new String("x") ("object")



and



typeof `x` ("string").



Those are two different things in JavaScript.






share|improve this answer


























  • actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

    – brk
    Apr 4 at 18:51





















4














JavaScript operator in only applicable to an Objects instances.



When you using constructor new String('abc') this will causing creating of a String object instance.



In other side, when you using only string literals or call function String('abc') without new it creates an string primitive. (like Number and Boolen)



Some behaviour of primitives and objects is differrent, look at this simple example's output:






console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.



But in it is not an method call, its language operator an in this case wrapping is not applied.



PS: Sorry for my english.






share|improve this answer

































    3














    typeof('test') == string (string literal)



    typof(new String('test')) == object (string object)



    you can't use in with a string literal.




    The in operator returns true if the specified property is in the specified object or its prototype chain.







    share|improve this answer































      2















      The in operator can only be used to check if a property is in an
      object. You can't search in strings, or in numbers, or other primitive
      types.




      The first example works and prints 'true' because length is a property of a string object.



      The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.






      share|improve this answer





















      • 1





        Notice, that let1.length works in the snippet.

        – Teemu
        Apr 4 at 18:53











      • Right. But let1 is a string, not an object in the second example.

        – VHS
        Apr 4 at 18:54






      • 1





        Umh ... the second example works as well.

        – Teemu
        Apr 4 at 19:01






      • 1





        Just run the second snippet, the first console.log shows 4.

        – Teemu
        Apr 4 at 19:04








      • 1





        Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

        – Teemu
        Apr 4 at 19:06














      Your Answer






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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      39














      In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).



      See Distinction between string primitives and String objects



      Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.




      You must specify an object on the right side of the in operator. For
      example, you can specify a string created with the String constructor,
      but you cannot specify a string literal.







      share|improve this answer





















      • 3





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        Apr 4 at 19:05











      • Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

        – ChrisM
        Apr 5 at 16:12


















      39














      In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).



      See Distinction between string primitives and String objects



      Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.




      You must specify an object on the right side of the in operator. For
      example, you can specify a string created with the String constructor,
      but you cannot specify a string literal.







      share|improve this answer





















      • 3





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        Apr 4 at 19:05











      • Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

        – ChrisM
        Apr 5 at 16:12
















      39












      39








      39







      In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).



      See Distinction between string primitives and String objects



      Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.




      You must specify an object on the right side of the in operator. For
      example, you can specify a string created with the String constructor,
      but you cannot specify a string literal.







      share|improve this answer















      In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).



      See Distinction between string primitives and String objects



      Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.




      You must specify an object on the right side of the in operator. For
      example, you can specify a string created with the String constructor,
      but you cannot specify a string literal.








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 4 at 19:16

























      answered Apr 4 at 18:56









      benvcbenvc

      6,9031928




      6,9031928








      • 3





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        Apr 4 at 19:05











      • Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

        – ChrisM
        Apr 5 at 16:12
















      • 3





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        Apr 4 at 19:05











      • Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

        – ChrisM
        Apr 5 at 16:12










      3




      3





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      Apr 4 at 19:05





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      Apr 4 at 19:05













      Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

      – ChrisM
      Apr 5 at 16:12







      Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.

      – ChrisM
      Apr 5 at 16:12















      10














      It throws an error because in is an operator for objects:




      prop in object




      but when you declare a string as `` (` string(template) literals) or "" '' (",' string literals) you don't create an object.



      Check



      typeof new String("x") ("object")



      and



      typeof `x` ("string").



      Those are two different things in JavaScript.






      share|improve this answer


























      • actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

        – brk
        Apr 4 at 18:51


















      10














      It throws an error because in is an operator for objects:




      prop in object




      but when you declare a string as `` (` string(template) literals) or "" '' (",' string literals) you don't create an object.



      Check



      typeof new String("x") ("object")



      and



      typeof `x` ("string").



      Those are two different things in JavaScript.






      share|improve this answer


























      • actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

        – brk
        Apr 4 at 18:51
















      10












      10








      10







      It throws an error because in is an operator for objects:




      prop in object




      but when you declare a string as `` (` string(template) literals) or "" '' (",' string literals) you don't create an object.



      Check



      typeof new String("x") ("object")



      and



      typeof `x` ("string").



      Those are two different things in JavaScript.






      share|improve this answer















      It throws an error because in is an operator for objects:




      prop in object




      but when you declare a string as `` (` string(template) literals) or "" '' (",' string literals) you don't create an object.



      Check



      typeof new String("x") ("object")



      and



      typeof `x` ("string").



      Those are two different things in JavaScript.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 5 at 18:06

























      answered Apr 4 at 18:49









      SkillGGSkillGG

      334112




      334112













      • actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

        – brk
        Apr 4 at 18:51





















      • actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

        – brk
        Apr 4 at 18:51



















      actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

      – brk
      Apr 4 at 18:51







      actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))

      – brk
      Apr 4 at 18:51













      4














      JavaScript operator in only applicable to an Objects instances.



      When you using constructor new String('abc') this will causing creating of a String object instance.



      In other side, when you using only string literals or call function String('abc') without new it creates an string primitive. (like Number and Boolen)



      Some behaviour of primitives and objects is differrent, look at this simple example's output:






      console.log(typeof (new String('ddd'))) // "object"
      console.log(typeof ('ddd')) // "string"
      console.log(eval('1 + 2')) // 3
      console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





      In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.



      But in it is not an method call, its language operator an in this case wrapping is not applied.



      PS: Sorry for my english.






      share|improve this answer






























        4














        JavaScript operator in only applicable to an Objects instances.



        When you using constructor new String('abc') this will causing creating of a String object instance.



        In other side, when you using only string literals or call function String('abc') without new it creates an string primitive. (like Number and Boolen)



        Some behaviour of primitives and objects is differrent, look at this simple example's output:






        console.log(typeof (new String('ddd'))) // "object"
        console.log(typeof ('ddd')) // "string"
        console.log(eval('1 + 2')) // 3
        console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





        In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.



        But in it is not an method call, its language operator an in this case wrapping is not applied.



        PS: Sorry for my english.






        share|improve this answer




























          4












          4








          4







          JavaScript operator in only applicable to an Objects instances.



          When you using constructor new String('abc') this will causing creating of a String object instance.



          In other side, when you using only string literals or call function String('abc') without new it creates an string primitive. (like Number and Boolen)



          Some behaviour of primitives and objects is differrent, look at this simple example's output:






          console.log(typeof (new String('ddd'))) // "object"
          console.log(typeof ('ddd')) // "string"
          console.log(eval('1 + 2')) // 3
          console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





          In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.



          But in it is not an method call, its language operator an in this case wrapping is not applied.



          PS: Sorry for my english.






          share|improve this answer















          JavaScript operator in only applicable to an Objects instances.



          When you using constructor new String('abc') this will causing creating of a String object instance.



          In other side, when you using only string literals or call function String('abc') without new it creates an string primitive. (like Number and Boolen)



          Some behaviour of primitives and objects is differrent, look at this simple example's output:






          console.log(typeof (new String('ddd'))) // "object"
          console.log(typeof ('ddd')) // "string"
          console.log(eval('1 + 2')) // 3
          console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





          In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.



          But in it is not an method call, its language operator an in this case wrapping is not applied.



          PS: Sorry for my english.






          console.log(typeof (new String('ddd'))) // "object"
          console.log(typeof ('ddd')) // "string"
          console.log(eval('1 + 2')) // 3
          console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}





          console.log(typeof (new String('ddd'))) // "object"
          console.log(typeof ('ddd')) // "string"
          console.log(eval('1 + 2')) // 3
          console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 5 at 12:06

























          answered Apr 4 at 18:49









          Stranger in the QStranger in the Q

          1,1061820




          1,1061820























              3














              typeof('test') == string (string literal)



              typof(new String('test')) == object (string object)



              you can't use in with a string literal.




              The in operator returns true if the specified property is in the specified object or its prototype chain.







              share|improve this answer




























                3














                typeof('test') == string (string literal)



                typof(new String('test')) == object (string object)



                you can't use in with a string literal.




                The in operator returns true if the specified property is in the specified object or its prototype chain.







                share|improve this answer


























                  3












                  3








                  3







                  typeof('test') == string (string literal)



                  typof(new String('test')) == object (string object)



                  you can't use in with a string literal.




                  The in operator returns true if the specified property is in the specified object or its prototype chain.







                  share|improve this answer













                  typeof('test') == string (string literal)



                  typof(new String('test')) == object (string object)



                  you can't use in with a string literal.




                  The in operator returns true if the specified property is in the specified object or its prototype chain.








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Apr 4 at 18:48









                  FedeScFedeSc

                  917923




                  917923























                      2















                      The in operator can only be used to check if a property is in an
                      object. You can't search in strings, or in numbers, or other primitive
                      types.




                      The first example works and prints 'true' because length is a property of a string object.



                      The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.






                      share|improve this answer





















                      • 1





                        Notice, that let1.length works in the snippet.

                        – Teemu
                        Apr 4 at 18:53











                      • Right. But let1 is a string, not an object in the second example.

                        – VHS
                        Apr 4 at 18:54






                      • 1





                        Umh ... the second example works as well.

                        – Teemu
                        Apr 4 at 19:01






                      • 1





                        Just run the second snippet, the first console.log shows 4.

                        – Teemu
                        Apr 4 at 19:04








                      • 1





                        Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                        – Teemu
                        Apr 4 at 19:06


















                      2















                      The in operator can only be used to check if a property is in an
                      object. You can't search in strings, or in numbers, or other primitive
                      types.




                      The first example works and prints 'true' because length is a property of a string object.



                      The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.






                      share|improve this answer





















                      • 1





                        Notice, that let1.length works in the snippet.

                        – Teemu
                        Apr 4 at 18:53











                      • Right. But let1 is a string, not an object in the second example.

                        – VHS
                        Apr 4 at 18:54






                      • 1





                        Umh ... the second example works as well.

                        – Teemu
                        Apr 4 at 19:01






                      • 1





                        Just run the second snippet, the first console.log shows 4.

                        – Teemu
                        Apr 4 at 19:04








                      • 1





                        Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                        – Teemu
                        Apr 4 at 19:06
















                      2












                      2








                      2








                      The in operator can only be used to check if a property is in an
                      object. You can't search in strings, or in numbers, or other primitive
                      types.




                      The first example works and prints 'true' because length is a property of a string object.



                      The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.






                      share|improve this answer
















                      The in operator can only be used to check if a property is in an
                      object. You can't search in strings, or in numbers, or other primitive
                      types.




                      The first example works and prints 'true' because length is a property of a string object.



                      The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited Apr 4 at 18:55

























                      answered Apr 4 at 18:53









                      VHSVHS

                      7,37631128




                      7,37631128








                      • 1





                        Notice, that let1.length works in the snippet.

                        – Teemu
                        Apr 4 at 18:53











                      • Right. But let1 is a string, not an object in the second example.

                        – VHS
                        Apr 4 at 18:54






                      • 1





                        Umh ... the second example works as well.

                        – Teemu
                        Apr 4 at 19:01






                      • 1





                        Just run the second snippet, the first console.log shows 4.

                        – Teemu
                        Apr 4 at 19:04








                      • 1





                        Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                        – Teemu
                        Apr 4 at 19:06
















                      • 1





                        Notice, that let1.length works in the snippet.

                        – Teemu
                        Apr 4 at 18:53











                      • Right. But let1 is a string, not an object in the second example.

                        – VHS
                        Apr 4 at 18:54






                      • 1





                        Umh ... the second example works as well.

                        – Teemu
                        Apr 4 at 19:01






                      • 1





                        Just run the second snippet, the first console.log shows 4.

                        – Teemu
                        Apr 4 at 19:04








                      • 1





                        Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                        – Teemu
                        Apr 4 at 19:06










                      1




                      1





                      Notice, that let1.length works in the snippet.

                      – Teemu
                      Apr 4 at 18:53





                      Notice, that let1.length works in the snippet.

                      – Teemu
                      Apr 4 at 18:53













                      Right. But let1 is a string, not an object in the second example.

                      – VHS
                      Apr 4 at 18:54





                      Right. But let1 is a string, not an object in the second example.

                      – VHS
                      Apr 4 at 18:54




                      1




                      1





                      Umh ... the second example works as well.

                      – Teemu
                      Apr 4 at 19:01





                      Umh ... the second example works as well.

                      – Teemu
                      Apr 4 at 19:01




                      1




                      1





                      Just run the second snippet, the first console.log shows 4.

                      – Teemu
                      Apr 4 at 19:04







                      Just run the second snippet, the first console.log shows 4.

                      – Teemu
                      Apr 4 at 19:04






                      1




                      1





                      Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                      – Teemu
                      Apr 4 at 19:06







                      Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs 4, the length of let1. Take a look at benvc's answer on this post.

                      – Teemu
                      Apr 4 at 19:06




















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