Why does indexOf not break?












10















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?






javascript angular typescript







share|improve this question

























  • 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
    59 mins ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    59 mins ago






  • 2





    @Jean-BaptisteYunès including null...

    – vlaz
    58 mins ago











  • @vlaz sarcasm ?

    – Florian
    44 mins ago






  • 1





    @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
    10 mins ago
















10















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?






javascript angular typescript







share|improve this question

























  • 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
    59 mins ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    59 mins ago






  • 2





    @Jean-BaptisteYunès including null...

    – vlaz
    58 mins ago











  • @vlaz sarcasm ?

    – Florian
    44 mins ago






  • 1





    @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
    10 mins ago














10












10








10








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?






javascript angular typescript







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?






javascript angular typescript




javascript angular typescript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 53 mins ago







De Wet van As

















asked 1 hour ago









De Wet van AsDe Wet van As

787




787













  • 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
    59 mins ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    59 mins ago






  • 2





    @Jean-BaptisteYunès including null...

    – vlaz
    58 mins ago











  • @vlaz sarcasm ?

    – Florian
    44 mins ago






  • 1





    @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
    10 mins 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
    59 mins ago






  • 1





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    59 mins ago






  • 2





    @Jean-BaptisteYunès including null...

    – vlaz
    58 mins ago











  • @vlaz sarcasm ?

    – Florian
    44 mins ago






  • 1





    @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
    10 mins 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
59 mins 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
59 mins ago




1




1





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
59 mins ago





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
59 mins ago




2




2





@Jean-BaptisteYunès including null...

– vlaz
58 mins ago





@Jean-BaptisteYunès including null...

– vlaz
58 mins ago













@vlaz sarcasm ?

– Florian
44 mins ago





@vlaz sarcasm ?

– Florian
44 mins ago




1




1





@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
10 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
10 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
    55 mins ago











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

    – Florian
    39 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
    5 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
      49 mins ago











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    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
      55 mins ago











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

      – Florian
      39 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
      55 mins ago











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

      – Florian
      39 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 59 mins 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
      55 mins ago











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

      – Florian
      39 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
      55 mins ago











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

      – Florian
      39 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
    55 mins 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
    55 mins ago













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

    – Florian
    39 mins ago





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

    – Florian
    39 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
      5 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
      5 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










    answered 59 mins ago









    0xc14m1z0xc14m1z

    1,409512




    1,409512








    • 1





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

      – Pavlo
      5 mins ago














    • 1





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

      – Pavlo
      5 mins ago








    1




    1





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

    – Pavlo
    5 mins ago





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

    – Pavlo
    5 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 59 mins 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
              49 mins 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
              49 mins 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 50 mins ago









            GibboKGibboK

            34.3k107317542




            34.3k107317542








            • 1





              What about TypeScript?

              – vlaz
              49 mins ago














            • 1





              What about TypeScript?

              – vlaz
              49 mins ago








            1




            1





            What about TypeScript?

            – vlaz
            49 mins ago





            What about TypeScript?

            – vlaz
            49 mins ago


















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