Why does accessing a property of indexOf still compile?












11















I made a typo in TypeScript which was picked up during code review.



I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).



I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.



Can anyone explain this?






typescript methods properties







share|improve this question




















  • 1





    What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

    – vlaz
    1 hour ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    1 hour ago






  • 2





    @DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

    – vlaz
    21 mins ago






  • 1





    @DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

    – Bergi
    7 mins ago
















11















I made a typo in TypeScript which was picked up during code review.



I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).



I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.



Can anyone explain this?






typescript methods properties







share|improve this question




















  • 1





    What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

    – vlaz
    1 hour ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    1 hour ago






  • 2





    @DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

    – vlaz
    21 mins ago






  • 1





    @DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

    – Bergi
    7 mins ago














11












11








11








I made a typo in TypeScript which was picked up during code review.



I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).



I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.



Can anyone explain this?






typescript methods properties







share|improve this question
















I made a typo in TypeScript which was picked up during code review.



I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).



I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.



Can anyone explain this?






typescript methods properties




typescript methods properties






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 10 mins ago









Bergi

366k58546872




366k58546872










asked 1 hour ago









De Wet van AsDe Wet van As

837




837








  • 1





    What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

    – vlaz
    1 hour ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    1 hour ago






  • 2





    @DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

    – vlaz
    21 mins ago






  • 1





    @DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

    – Bergi
    7 mins ago














  • 1





    What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

    – vlaz
    1 hour ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    1 hour ago






  • 2





    @DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

    – vlaz
    21 mins ago






  • 1





    @DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

    – Bergi
    7 mins ago








1




1





What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

– vlaz
1 hour ago





What did you assign the result of? Because what you wrote is valid you are taking the someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.

– vlaz
1 hour ago




1




1





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
1 hour ago





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
1 hour ago




2




2





@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

– vlaz
21 mins ago





@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.

– vlaz
21 mins ago




1




1





@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

– Bergi
7 mins ago





@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.

– Bergi
7 mins ago












4 Answers
4






active

oldest

votes


















15














Quite easy.



someArray.indexOf you know that this is a function, which is also an object and can have properties.



By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.



Of course, it is not defined on the indexOf function, so it returns undefined.



Quick example that illustrates the syntax and the fact that a function can have properties ;) : 






const array = ;

array.indexOf['anyValue'] = 'test';

console.log(array.indexOf.anyValue);








share|improve this answer



















  • 3





    I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

    – vlaz
    1 hour ago











  • @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

    – Florian
    50 mins ago



















3














array.indexOf is a function.



Functions are objects.



You were accessing the someObject property of the array.indexOf function.



You would have got undefined.






const array = [1, 2, 3]

const someObject = 'asdasd'



console.log(array.indexOf[someObject])

// undefined








share|improve this answer





















  • 1





    Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

    – Pavlo
    17 mins ago











  • You’re right! I fixed the typo in my answer. Thanks

    – 0xc14m1z
    10 mins ago



















0














In short: Because that is the way the language is designed.



JavaScript evaluates an attempt to access a property that doesn't exist as undefined.



It doesn't raise an exception.






share|improve this answer































    0














    Functions in JavaScript are first class objects.



    When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.






    share|improve this answer



















    • 1





      What about TypeScript?

      – vlaz
      1 hour ago











    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54254396%2fwhy-does-accessing-a-property-of-indexof-still-compile%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15














    Quite easy.



    someArray.indexOf you know that this is a function, which is also an object and can have properties.



    By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.



    Of course, it is not defined on the indexOf function, so it returns undefined.



    Quick example that illustrates the syntax and the fact that a function can have properties ;) : 






    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);








    share|improve this answer



















    • 3





      I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

      – vlaz
      1 hour ago











    • @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

      – Florian
      50 mins ago
















    15














    Quite easy.



    someArray.indexOf you know that this is a function, which is also an object and can have properties.



    By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.



    Of course, it is not defined on the indexOf function, so it returns undefined.



    Quick example that illustrates the syntax and the fact that a function can have properties ;) : 






    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);








    share|improve this answer



















    • 3





      I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

      – vlaz
      1 hour ago











    • @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

      – Florian
      50 mins ago














    15












    15








    15







    Quite easy.



    someArray.indexOf you know that this is a function, which is also an object and can have properties.



    By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.



    Of course, it is not defined on the indexOf function, so it returns undefined.



    Quick example that illustrates the syntax and the fact that a function can have properties ;) : 






    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);








    share|improve this answer













    Quite easy.



    someArray.indexOf you know that this is a function, which is also an object and can have properties.



    By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.



    Of course, it is not defined on the indexOf function, so it returns undefined.



    Quick example that illustrates the syntax and the fact that a function can have properties ;) : 






    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);








    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);





    const array = ;
    
array.indexOf['anyValue'] = 'test';
    
console.log(array.indexOf.anyValue);






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    sjahansjahan

    3,2051827




    3,2051827








    • 3





      I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

      – vlaz
      1 hour ago











    • @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

      – Florian
      50 mins ago














    • 3





      I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

      – vlaz
      1 hour ago











    • @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

      – Florian
      50 mins ago








    3




    3





    I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

    – vlaz
    1 hour ago





    I think answers here are missing the TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.

    – vlaz
    1 hour ago













    @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

    – Florian
    50 mins ago





    @vlaz +1. sjahan gives OP a quick explanation of the undefined result but the main question remains...

    – Florian
    50 mins ago













    3














    array.indexOf is a function.



    Functions are objects.



    You were accessing the someObject property of the array.indexOf function.



    You would have got undefined.






    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined








    share|improve this answer





















    • 1





      Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

      – Pavlo
      17 mins ago











    • You’re right! I fixed the typo in my answer. Thanks

      – 0xc14m1z
      10 mins ago
















    3














    array.indexOf is a function.



    Functions are objects.



    You were accessing the someObject property of the array.indexOf function.



    You would have got undefined.






    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined








    share|improve this answer





















    • 1





      Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

      – Pavlo
      17 mins ago











    • You’re right! I fixed the typo in my answer. Thanks

      – 0xc14m1z
      10 mins ago














    3












    3








    3







    array.indexOf is a function.



    Functions are objects.



    You were accessing the someObject property of the array.indexOf function.



    You would have got undefined.






    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined








    share|improve this answer















    array.indexOf is a function.



    Functions are objects.



    You were accessing the someObject property of the array.indexOf function.



    You would have got undefined.






    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined








    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined





    const array = [1, 2, 3]
    
const someObject = 'asdasd'
    

    
console.log(array.indexOf[someObject])
    
// undefined






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 11 mins ago

























    answered 1 hour ago









    0xc14m1z0xc14m1z

    1,409512




    1,409512








    • 1





      Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

      – Pavlo
      17 mins ago











    • You’re right! I fixed the typo in my answer. Thanks

      – 0xc14m1z
      10 mins ago














    • 1





      Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

      – Pavlo
      17 mins ago











    • You’re right! I fixed the typo in my answer. Thanks

      – 0xc14m1z
      10 mins ago








    1




    1





    Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

    – Pavlo
    17 mins ago





    Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.

    – Pavlo
    17 mins ago













    You’re right! I fixed the typo in my answer. Thanks

    – 0xc14m1z
    10 mins ago





    You’re right! I fixed the typo in my answer. Thanks

    – 0xc14m1z
    10 mins ago











    0














    In short: Because that is the way the language is designed.



    JavaScript evaluates an attempt to access a property that doesn't exist as undefined.



    It doesn't raise an exception.






    share|improve this answer




























      0














      In short: Because that is the way the language is designed.



      JavaScript evaluates an attempt to access a property that doesn't exist as undefined.



      It doesn't raise an exception.






      share|improve this answer


























        0












        0








        0







        In short: Because that is the way the language is designed.



        JavaScript evaluates an attempt to access a property that doesn't exist as undefined.



        It doesn't raise an exception.






        share|improve this answer













        In short: Because that is the way the language is designed.



        JavaScript evaluates an attempt to access a property that doesn't exist as undefined.



        It doesn't raise an exception.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        QuentinQuentin

        642k718661036




        642k718661036























            0














            Functions in JavaScript are first class objects.



            When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.






            share|improve this answer



















            • 1





              What about TypeScript?

              – vlaz
              1 hour ago
















            0














            Functions in JavaScript are first class objects.



            When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.






            share|improve this answer



















            • 1





              What about TypeScript?

              – vlaz
              1 hour ago














            0












            0








            0







            Functions in JavaScript are first class objects.



            When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.






            share|improve this answer













            Functions in JavaScript are first class objects.



            When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 1 hour ago









            GibboKGibboK

            34.3k107317542




            34.3k107317542








            • 1





              What about TypeScript?

              – vlaz
              1 hour ago














            • 1





              What about TypeScript?

              – vlaz
              1 hour ago








            1




            1





            What about TypeScript?

            – vlaz
            1 hour ago





            What about TypeScript?

            – vlaz
            1 hour ago


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54254396%2fwhy-does-accessing-a-property-of-indexof-still-compile%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Statuo de Libereco

            Image
            Koordinatoj: 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Statuo de Libereco angle: Statue of Liberty statuo Statuo de Libereco de antaŭe Oficiala nomo: Liberty Enlightening the World Lando  Usono Ŝtato Novjorkio Regiono Novjorko Insulo Insulo de Libereco Situo Statuo de Libereco  - koordinatoj 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Alteco de statuo 46 m  - alteco kun soklo 93 m Pezo 205 000 kg (451 948 lb) Materialoj bronzo, ŝtalo Proponinto Gustave Eiffel  - skulptisto Frédéric Auguste Bartholdi  - proponinto de soklo Richard Morris Hunt  - kapo laŭ modelo de Isabella Eugenie Boyer Transdonita 4-an de julio 1886 Por publiko alirebla por publiko Horzono OT (UTC-5)  - somera tempo OT (UTC-4) Loko
            Read more

            Tanganjiko

            Image
            Lago Tanganjiko Situo centra Afriko Ĉefaj fontoj Kalambo, Lufuko, Malagarasi kaj Ruzizi Ĉefaj elfluoj Lukugo Setlejoj Bujumbura, Kalemie Datenoj Koordinatoj 6°  S , 30°  O -6.3827777777778 29.615 782 Koordinatoj: 6°  S , 30°  O DEC Supermara alteco 782 m Surfaca areo 32 893 km² f5 Maksimuma longo 673 km f6 Maksimuma larĝo 72 km f7 Akva volumeno 18.900 km³ dep1 f8 Maksimuma profundo 1470 m f10 Averaĝa profundo 570 m f11 v   •   d   •   r Mapo de la lago Tanganjiko estas lago en orienta Afriko, inter Burundo, Kongo Kinŝasa, Tanzanio kaj Zambio. Ĝi apartenas al la Granda Rifto kaj estas unu el la Afrikaj Grandaj Lagoj. Tanzanio (46%) kaj DRK (40%) havas la majoritaton de la lago. Ĝi havas longon de 673 km, areon de ĉ. 32.900 km² kaj maksimuman profundon taksatan je 1470 metroj; pro tiu ĉi granda profundo ankaŭ ĝia volumeno estas sufiĉe granda, 18.900 km³. Oni konsi
            Read more

            Liste der Baudenkmäler in Enneberg

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