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





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7















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);








javascript string







share|improve this question

























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

    – Teemu
    7 hours ago













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

    – Bergi
    7 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    7 hours ago


















7















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);








javascript string







share|improve this question

























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

    – Teemu
    7 hours ago













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

    – Bergi
    7 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    7 hours ago














7












7








7








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);








javascript string







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




javascript string






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Boann

37.4k1290122




37.4k1290122










asked 7 hours ago









brkbrk

29.8k32244




29.8k32244













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

    – Teemu
    7 hours ago













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

    – Bergi
    7 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    7 hours ago



















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

    – Teemu
    7 hours ago













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

    – Bergi
    7 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    7 hours ago

















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

– Teemu
7 hours ago







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

– Teemu
7 hours ago















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

– Bergi
7 hours ago





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

– Bergi
7 hours ago













@Bergi Well, that explains a lot.

– Teemu
7 hours ago





@Bergi Well, that explains a lot.

– Teemu
7 hours ago












5 Answers
5






active

oldest

votes


















10














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





















  • 2





    This is the shortest and most concise correct answer shown.

    – Scott Marcus
    7 hours ago



















6














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




prop in object




but when you declare a string as `` (` string 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
    7 hours ago





















3














Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.






console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))








share|improve this answer

































    2














    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































      1















      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


























      • Notice, that let1.length works in the snippet.

        – Teemu
        7 hours ago











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

        – VHS
        7 hours ago











      • Umh ... the second example works as well.

        – Teemu
        7 hours ago











      • The second example wouldn't work because let1 is not an object.

        – VHS
        7 hours ago











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

        – Teemu
        7 hours ago














      Your Answer






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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10














      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





















      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        7 hours ago
















      10














      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





















      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        7 hours ago














      10












      10








      10







      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 6 hours ago

























      answered 7 hours ago









      benvcbenvc

      6,6081828




      6,6081828








      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        7 hours ago














      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        7 hours ago








      2




      2





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      7 hours ago





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      7 hours ago













      6














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




      prop in object




      but when you declare a string as `` (` string 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
        7 hours ago


















      6














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




      prop in object




      but when you declare a string as `` (` string 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
        7 hours ago
















      6












      6








      6







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




      prop in object




      but when you declare a string as `` (` string 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 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 4 hours ago









      Boann

      37.4k1290122




      37.4k1290122










      answered 7 hours ago









      SkillGGSkillGG

      1839




      1839













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

        – brk
        7 hours ago





















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

        – brk
        7 hours ago



















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

      – brk
      7 hours ago







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

      – brk
      7 hours ago













      3














      Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.






      console.log(typeof (new String('ddd')))
      
console.log(typeof ('ddd'))








      share|improve this answer






























        3














        Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.






        console.log(typeof (new String('ddd')))
        
console.log(typeof ('ddd'))








        share|improve this answer




























          3












          3








          3







          Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.






          console.log(typeof (new String('ddd')))
          
console.log(typeof ('ddd'))








          share|improve this answer















          Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.






          console.log(typeof (new String('ddd')))
          
console.log(typeof ('ddd'))








          console.log(typeof (new String('ddd')))
          
console.log(typeof ('ddd'))





          console.log(typeof (new String('ddd')))
          
console.log(typeof ('ddd'))






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Stranger in the QStranger in the Q

          7161618




          7161618























              2














              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














                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












                  2








                  2







                  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 7 hours ago









                  FedeScFedeSc

                  877923




                  877923























                      1















                      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


























                      • Notice, that let1.length works in the snippet.

                        – Teemu
                        7 hours ago











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

                        – VHS
                        7 hours ago











                      • Umh ... the second example works as well.

                        – Teemu
                        7 hours ago











                      • The second example wouldn't work because let1 is not an object.

                        – VHS
                        7 hours ago











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

                        – Teemu
                        7 hours ago


















                      1















                      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


























                      • Notice, that let1.length works in the snippet.

                        – Teemu
                        7 hours ago











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

                        – VHS
                        7 hours ago











                      • Umh ... the second example works as well.

                        – Teemu
                        7 hours ago











                      • The second example wouldn't work because let1 is not an object.

                        – VHS
                        7 hours ago











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

                        – Teemu
                        7 hours ago
















                      1












                      1








                      1








                      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 7 hours ago

























                      answered 7 hours ago









                      VHSVHS

                      7,22431128




                      7,22431128













                      • Notice, that let1.length works in the snippet.

                        – Teemu
                        7 hours ago











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

                        – VHS
                        7 hours ago











                      • Umh ... the second example works as well.

                        – Teemu
                        7 hours ago











                      • The second example wouldn't work because let1 is not an object.

                        – VHS
                        7 hours ago











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

                        – Teemu
                        7 hours ago





















                      • Notice, that let1.length works in the snippet.

                        – Teemu
                        7 hours ago











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

                        – VHS
                        7 hours ago











                      • Umh ... the second example works as well.

                        – Teemu
                        7 hours ago











                      • The second example wouldn't work because let1 is not an object.

                        – VHS
                        7 hours ago











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

                        – Teemu
                        7 hours ago



















                      Notice, that let1.length works in the snippet.

                      – Teemu
                      7 hours ago





                      Notice, that let1.length works in the snippet.

                      – Teemu
                      7 hours ago













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

                      – VHS
                      7 hours ago





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

                      – VHS
                      7 hours ago













                      Umh ... the second example works as well.

                      – Teemu
                      7 hours ago





                      Umh ... the second example works as well.

                      – Teemu
                      7 hours ago













                      The second example wouldn't work because let1 is not an object.

                      – VHS
                      7 hours ago





                      The second example wouldn't work because let1 is not an object.

                      – VHS
                      7 hours ago













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

                      – Teemu
                      7 hours ago







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

                      – Teemu
                      7 hours ago




















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                      Liste der Baudenkmäler in Enneberg

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                      Die Liste der Baudenkmäler in Enneberg (ladinisch Mareo , italienisch Marebbe ) enthält die 36 als Baudenkmäler ausgewiesenen Objekte auf dem Gebiet der Gemeinde Enneberg in Südtirol. Basis ist das im Internet einsehbare offizielle Verzeichnis der Baudenkmäler in Südtirol. Dabei kann es sich beispielsweise um Sakralbauten, Wohnhäuser, Bauernhöfe und Adelsansitze handeln. Die Reihenfolge in dieser Liste orientiert sich an der Bezeichnung, alternativ ist sie auch nach der Adresse oder dem Datum der Unterschutzstellung sortierbar. Liste | Foto   Bezeichnung Standort Eintragung Beschreibung Metadaten ja Altes Gasthaus Janesch ID: 14567 46° 42′ 54″ N, 11° 52′ 42″ O 76882557800 0 8. Sep. 1986 (BLR-LAB 4974) Ländliches Wohnhaus/Bauernhof KG:  Welschellen Bauparzelle: 391 ja Asch (Prack) ID: 14540 46° 43′ 12″ N, 11° 54′ 35″ O 75496485300 20. Nov. 1950 (MD) Ansitz KG:  Enneberg Bauparzelle: 38 ja Biccia mit Stade
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