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Why 'in' operator is throwing error instead of logging false with string literal


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”?Why String Pool behaves differently for literals and variables?






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6















As per mdn the in operator returns true if the property exist and according to that the first example is logging true, but when creating a string literals why it is throwing 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
    2 hours ago












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

    – Bergi
    2 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    2 hours ago

















6















As per mdn the in operator returns true if the property exist and according to that the first example is logging true, but when creating a string literals why it is throwing 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
    2 hours ago












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

    – Bergi
    2 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    2 hours ago













6












6








6








As per mdn the in operator returns true if the property exist and according to that the first example is logging true, but when creating a string literals why it is throwing 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 exist and according to that the first example is logging true, but when creating a string literals why it is throwing 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 2 hours ago









benvc

6,5481827




6,5481827










asked 2 hours ago









brkbrk

29.8k32244




29.8k32244












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

    – Teemu
    2 hours ago












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

    – Bergi
    2 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    2 hours ago

















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

    – Teemu
    2 hours ago












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

    – Bergi
    2 hours ago











  • @Bergi Well, that explains a lot.

    – Teemu
    2 hours ago
















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

– Teemu
2 hours ago






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

– Teemu
2 hours ago














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

– Bergi
2 hours ago





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

– Bergi
2 hours ago













@Bergi Well, that explains a lot.

– Teemu
2 hours ago





@Bergi Well, that explains a lot.

– Teemu
2 hours ago












5 Answers
5






active

oldest

votes


















5














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




prop in object




but when you declare 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
    2 hours ago



















4














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
    2 hours ago


















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














    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
































      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
        2 hours ago











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

        – VHS
        2 hours ago











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

        – Teemu
        2 hours ago











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

        – VHS
        2 hours ago











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

        – Teemu
        2 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









      5














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




      prop in object




      but when you declare 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
        2 hours ago
















      5














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




      prop in object




      but when you declare 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
        2 hours ago














      5












      5








      5







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




      prop in object




      but when you declare 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 error, because in is an operator for objects,




      prop in object




      but when you declare 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 2 hours ago

























      answered 2 hours ago









      SkillGGSkillGG

      1739




      1739












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

        – brk
        2 hours ago


















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

        – brk
        2 hours ago

















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

      – brk
      2 hours ago






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

      – brk
      2 hours ago














      4














      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
        2 hours ago















      4














      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
        2 hours ago













      4












      4








      4







      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 2 hours ago

























      answered 2 hours ago









      benvcbenvc

      6,5481827




      6,5481827







      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        2 hours ago












      • 2





        This is the shortest and most concise correct answer shown.

        – Scott Marcus
        2 hours ago







      2




      2





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      2 hours ago





      This is the shortest and most concise correct answer shown.

      – Scott Marcus
      2 hours ago











      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 2 hours ago









          FedeScFedeSc

          877923




          877923





















              2














              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














                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












                  2








                  2







                  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 2 hours ago

























                  answered 2 hours ago









                  Stranger in the QStranger in the Q

                  7061618




                  7061618





















                      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
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 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
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 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 2 hours ago

























                      answered 2 hours ago









                      VHSVHS

                      7,22431128




                      7,22431128












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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago


















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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago











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

                        – VHS
                        2 hours ago











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

                        – Teemu
                        2 hours ago

















                      Notice, that let1.length works in the snippet.

                      – Teemu
                      2 hours ago





                      Notice, that let1.length works in the snippet.

                      – Teemu
                      2 hours ago













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

                      – VHS
                      2 hours ago





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

                      – VHS
                      2 hours ago













                      Umh ... the second example works as well.

                      – Teemu
                      2 hours ago





                      Umh ... the second example works as well.

                      – Teemu
                      2 hours ago













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

                      – VHS
                      2 hours ago





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

                      – VHS
                      2 hours ago













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

                      – Teemu
                      2 hours ago






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

                      – Teemu
                      2 hours ago


















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                      Personnalité américaine du monde des affairesPersonnalité féminine françaiseNaissance en décembre 1841Naissance à ParisDécès en mai 1904Décès à 62 ansDécès à Paris 17 décembre1841Paris12 mai1904Isaac Merritt SingerSingerNew York1863guerre civile américaineParisAngleterreguerre de 1870DevonOldway Mansion (en) PaigntonWinnarettaEdmond de Polignacduchesse DecazesdollarsVictor ReubsaetParisHumbert I er d'Italieviolon de StradivariusStatue de la LibertéBartholdi Isabella Eugénie Boyer.mw-parser-output .entete.humainbackground-image:url("//upload.wikimedia.org/wikipedia/commons/8/82/Picto_infobox_manwoman.png") Isabella Eugénie Boyer. Titre de noblesse Duchesse Biographie Naissance 17 décembre 1841 Paris Décès 12 mai 1904 (à 62 ans) Paris Sépulture Cimetière de Passy Nationalité Française Domicile Oldway Mansion ( en ) Activité Mannequin Con...
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