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


Why can't I use switch statement on a String?Python join: why is it string.join(list) instead of list.join(string)?Multiline String Literal in C#Why does comparing strings using either '==' or 'is' sometimes produce a different result?How to initialize an array's length in javascript?How can I print literal curly-brace characters in python string and also use .format on it?Why does ++[[]][+[]]+[+[]] return the string “10”?Why is char[] preferred over String for passwords?Why does this code using random strings print “hello world”?jQuery.inArray(), how to use it right?






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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
    4 hours ago












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

    – Bergi
    4 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    4 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
    4 hours ago












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

    – Bergi
    4 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    4 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 1 hour ago









Boann

37.4k1290122




37.4k1290122










asked 4 hours ago









brkbrk

29.8k32244




29.8k32244












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

    – Teemu
    4 hours ago












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

    – Bergi
    4 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    4 hours ago

















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

    – Teemu
    4 hours ago












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

    – Bergi
    4 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    4 hours ago
















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

– Teemu
4 hours ago






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

– Teemu
4 hours ago














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

– Bergi
4 hours ago





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

– Bergi
4 hours ago













@Bergi Well, that explains a lot.

– Teemu
4 hours ago





@Bergi Well, that explains a lot.

– Teemu
4 hours ago












5 Answers
5






active

oldest

votes


















6














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
    4 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
    4 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
        4 hours ago











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

        – VHS
        4 hours ago











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

        – Teemu
        4 hours ago











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

        – VHS
        4 hours ago











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

        – Teemu
        4 hours ago












      Your Answer






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      5 Answers
      5






      active

      oldest

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      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      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
        4 hours ago















      6














      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
        4 hours ago













      6












      6








      6







      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 4 hours ago

























      answered 4 hours ago









      benvcbenvc

      6,5681827




      6,5681827







      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        4 hours ago












      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        4 hours ago







      2




      2





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      4 hours ago





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      4 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
        4 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
        4 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 1 hour ago









      Boann

      37.4k1290122




      37.4k1290122










      answered 4 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
        4 hours ago


















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

        – brk
        4 hours ago

















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

      – brk
      4 hours ago






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

      – brk
      4 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 4 hours ago

























          answered 4 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 4 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
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 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
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 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 4 hours ago

























                      answered 4 hours ago









                      VHSVHS

                      7,22431128




                      7,22431128












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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago


















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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago











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

                        – VHS
                        4 hours ago











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

                        – Teemu
                        4 hours ago

















                      Notice, that let1.length works in the snippet.

                      – Teemu
                      4 hours ago





                      Notice, that let1.length works in the snippet.

                      – Teemu
                      4 hours ago













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

                      – VHS
                      4 hours ago





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

                      – VHS
                      4 hours ago













                      Umh ... the second example works as well.

                      – Teemu
                      4 hours ago





                      Umh ... the second example works as well.

                      – Teemu
                      4 hours ago













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

                      – VHS
                      4 hours ago





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

                      – VHS
                      4 hours ago













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

                      – Teemu
                      4 hours ago






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

                      – Teemu
                      4 hours ago


















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                      Städer i Lillpolens vojvodskapStadskommuner i Lillpolens vojvodskap stadPolenKatowiceKraków2006kronlandetGalizienSołasWeichselösterrikarepreussareAuschwitzZatorpiastiska kungahusettyska förbundetkoncentrationslägretAuschwitzandra världskrigetförintelselägerFörintelsenmuseumOświęcims synagoga Koordinater: 50°2′18″N 19°13′17″Ö  /  50.03833°N 19.22139°Ö  / 50.03833; 19.22139 Oświęcim Från Wikipedia Hoppa till navigering Hoppa till sök .mw-parser-output .infoboxborder:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em.mw-parser-output .infobox td,.mw-parser-output .infobox thvertical-align:top;padding:0 .2em.mw-parser-output .infobox captionfont-size:larger.mw-parser-output .inf...
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                      Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

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                      Valle di Casies tedescocomune italianoprovincia autonoma di BolzanoTrentino-Alto AdigeValVal Pusteriario CasiesMonguelfoAustriatoponimo117811891299tedescoXII secolo1929municipio1985ted.cappuccinoJoachim HaspingerAndreas HoferbavaresifrancesinapoleonicibronzoOthmar Winklerbunkersbarramento Planca di SopraVallo Alpino in Alto Adigesci di fondoGran fondo della Val Casiesskiliftpiste da slittino naturale Valle di Casies Da Wikipedia, l'enciclopedia libera. Jump to navigation Jump to search Valle di Casies comune ( IT ) Valle di Casies ( DE ) Gsies Una veduta della vallata Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Kurti Taschler (SVP) dal 10-5-2015 (1º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°46′N 12°11′E  /  46.766667°N 12.183333°E 46.766667; 12.183333  ( Valle di Casies ) Coordinate: 46°46...
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                      Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager

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