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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
    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 3 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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                      कुँवर स्रोत दिक्चालन सूची"कुँवर""राणा कुँवरके वंशावली"

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                      पुष्टि की ज़रूरत वाले "सम्बंधित जाति समूह" क्षत्रियक्षेत्रीकर्णालीक्षेत्रीनेपाली भाषामेंकाजीबिष्टपाण्डेबस्नेतथापावत्सपृथ्वीनारायण शाहकेबलभद्र कुंवरभीमसेन थापाकेजंगबहादुर राणाराणा कुँवर/कँवर (गोरखाली) उपर: रामकृष्ण कुँवर बीच:बलभद्र कुंवर  · बालनरसिंह कुँवर तल:जंगबहादुर कुँवर राणा  · बमबहादुर कुँवर राणा कुल जनसंख्या ८२,१४९ नेपालमे (राणा कुँवरको छोडके) [1] [2] ख़ास आवास क्षेत्र   नेपाल भाषाएँ नेपाली भाषा धर्म हिन्दु धर्म और मष्टो धर्म अन्य सम्बंधित समूह पाँडे, थापा, बस्नेत, बिष्ट कुँवर वा कँवर नेपाली क्षत्रिय वर्णके क्षेत्री जातिके पारिवारिक नाम है। इस परिवारकी उत्पत्ति नेपालके कर्णाली प्रदेशमे अन्य क्षेत्री जातिके साथ हुआ था। कुँवरका अर्थ नेपाली भाषामें कुमार है। यिनका भारतीय राजपुत और मराठा राजाओं से कुछ वैवाहिक सम्बन्ध है। भारतमे पाएजानेवाले कुँवर अलग वंशजके है। कुँवर पाँच पुस्त्यौनी काजी परिवार मध्ये एक है। अन्य चार परिवार बिष्ट, पाण्डे, बस्नेत और थापा है। [3] इनके गोत्र वत्स गोत्र है। [4] रामकृष्ण कुँवर पृथ्वीनारायण शाहके सेनामे सरदार थे। बलभद्र कुंवर भीमसेन थापाके
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                      शेव्रोले वोल्ट अनुक्रम इतिहास इन्हे भी देखें चित्र दीर्घा संदर्भ दिक्चालन सूची

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                      कार अमेरिकीऑस्ट्रेलियान्यूजीलैंडयूनाइटेड किंगडमयूरोपइंजनलीड-एसिड बैटरी शेव्रोले वोल्ट अवलोकन निर्माता जनरल मोटर्स निर्माण 2011 शेव्रोले वोल्ट एक अमेरिकी कार निर्माता, जनरल मोटर्स द्वारा निर्मित एक प्लग-इन हाइब्रिड कार है. इसे चीन में बुइक वेलाइट 5 के रूप में, [1] ऑस्ट्रेलिया और न्यूजीलैंड में होल्डन वोल्ट , और एक अलग बनावट के साथ वॉक्सहॉल एम्पेरा के रूप में, यूनाइटेड किंगडम और यूरोप के बाकी हिस्सों में ओपल अम्पेरा के नाम से पुनर्निर्मित रूपों में भी विपणन किया (बेचा) जाता है। इस कि बिक्री यूएस में मध्य दिसंबर २०१० से, और यूरोप में २०११ से शुरु हुई। [2] वोल्ट एक शुद्ध बैटरी इलेक्ट्रिक वाहन के रूप में काम करती है जब तक इसकी बैटरी क्षमता पूर्ण चार्ज से पूर्व निर्धारित सीमा तक गिर ना जाए। वहां से इसका आंतरिक दहन इंजन आवश्यकतानुसार वाहन की सीमा का विस्तार करने के लिए एक विद्युत जनरेटर को शक्ति देता है। जब इंजन चलता है, तो ऊर्जा दक्षता में सुधार करने के लिए, यह समय-समय पर यांत्रिक रूप से (एक क्लच द्वारा) एक ग्रह गियर सेट से, और इसलिए आउटपुट ड्राइव धुरी से जुड़ जाता है । वोल्ट की रिजे
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                      चैत्य भूमि चित्र दीर्घा सन्दर्भ बाहरी कडियाँ दिक्चालन सूची"Chaitya Bhoomi""Chaitya Bhoomi: Statue of Equality in India""Dadar Chaitya Bhoomi: Statue of Equality in India""Ambedkar memorial: Centre okays transfer of Indu Mill land"चैत्यभमि

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                      मुम्बईबौद्ध तीर्थ स्थलभीमराव आंबेडकर को समर्पित स्मारक मुंबईदादरभारतीय संविधानडॉ भीमराव आंबेडकरबौद्ध धर्मदिल्लीमहापरिनिर्वाणमुंबईभदन्त आनन्द कौशल्यायनअरबी समुद्र चैत्यभूमि चैत्यभूमि मुंबई के दादर स्थित भारतीय संविधान के निर्माता भारतरत्न डॉ भीमराव आंबेडकर जी की समाधि स्थली और बौद्ध धर्म के लोगो का आस्था का केंद्र हैं। दादर के समुद्र तट पर चैत्य भूमि स्थित है। [1] [2] [3] [4] दिल्ली में अपने निवास में ६ दिसंबर १९५६ को आंबेडकर का महापरिनिर्वाण हुआ और उनका पार्थिव मुंबई में लाया गया, चैत्यभूमि में ७ दिसंबर १९५६ को भदन्त आनन्द कौशल्यायन ने आंबेडकर का बौद्ध परम्परा के अनुसार दाह संस्कार किया था। आंबेडकर का दाह संस्कार के पहले उन्हें साक्षी रख उनके १०,००,००० अनुयायिओं ने बौद्ध धम्म की दीक्षा ली और यह दीक्षा उन्हें भदन्त आनन्द कौशल्यायन ने दी थी। चित्र दीर्घा चैत्यभूमि स्तूप अशोकस्तंभ अशोकस्तंभ व प्रवेशद्वार (अंदर से) प्रवेशद्वार (अंदर से) चैत्यभूमि परिसर, आगे अशोकस्तंभ और अरबी समुद्र चैत्यभूम
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