Why does accessing a property of indexOf still compile?












20















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?










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
    2 days ago






  • 11





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    2 days ago






  • 7





    @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
    2 days ago











  • @vlaz it was in fact a complex object with multiple properties.

    – De Wet van As
    2 days ago
















20















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?










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
    2 days ago






  • 11





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    2 days ago






  • 7





    @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
    2 days ago











  • @vlaz it was in fact a complex object with multiple properties.

    – De Wet van As
    2 days ago














20












20








20


1






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?










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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago









Bergi

366k58546872




366k58546872










asked 2 days ago









De Wet van AsDe Wet van As

1318




1318








  • 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
    2 days ago






  • 11





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    2 days ago






  • 7





    @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
    2 days ago











  • @vlaz it was in fact a complex object with multiple properties.

    – De Wet van As
    2 days 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
    2 days ago






  • 11





    Welcome to javascript, where everything is an object!

    – Jean-Baptiste Yunès
    2 days ago






  • 7





    @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
    2 days ago











  • @vlaz it was in fact a complex object with multiple properties.

    – De Wet van As
    2 days 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
2 days 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
2 days ago




11




11





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
2 days ago





Welcome to javascript, where everything is an object!

– Jean-Baptiste Yunès
2 days ago




7




7





@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
2 days 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
2 days ago













@vlaz it was in fact a complex object with multiple properties.

– De Wet van As
2 days ago





@vlaz it was in fact a complex object with multiple properties.

– De Wet van As
2 days ago












4 Answers
4






active

oldest

votes


















28














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);





EDIT



Here is an attempt of an answer for the TypeScript side of the question.



As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:




  • 'Statically': obj.property

  • 'Dynamically': obj['property']


By using the 'static' way to access a property, of course, TypeScript will raise an error!



But with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.



That's why it will be implicitly marked as any.



As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny, please refer to his answer for more details about this!



Hoping this helps ;)






share|improve this answer





















  • 4





    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
    2 days ago











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

    – Florian
    2 days ago






  • 1





    "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

    – vlaz
    2 days ago











  • @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

    – sjahan
    2 days ago



















15














It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:



noImplicitAny enabled



The reason is that an element access expression returns an object typed as any when the type has no index signature defined (this is an implicit any).



enter image description here



So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.






share|improve this answer

































    4














    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
      2 days ago











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

      – 0xc14m1z
      2 days ago



















    1














    The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject) function.



    What happened when you used square braces, someArray.indexOf[someObject], was that the JS runtime first converted your object someObject into a string by calling the function someObject.toString, which most likely returned "[object object]". Then the someArray.indexOf object was queried for the key "[object object]" which wasn't present, returning undefined. As far as Typescript goes, this is completely fine.



    David Sherret pointed out that --noImplicitAny would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.






    share|improve this answer























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      28














      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);





      EDIT



      Here is an attempt of an answer for the TypeScript side of the question.



      As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:




      • 'Statically': obj.property

      • 'Dynamically': obj['property']


      By using the 'static' way to access a property, of course, TypeScript will raise an error!



      But with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.



      That's why it will be implicitly marked as any.



      As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny, please refer to his answer for more details about this!



      Hoping this helps ;)






      share|improve this answer





















      • 4





        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
        2 days ago











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

        – Florian
        2 days ago






      • 1





        "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

        – vlaz
        2 days ago











      • @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

        – sjahan
        2 days ago
















      28














      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);





      EDIT



      Here is an attempt of an answer for the TypeScript side of the question.



      As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:




      • 'Statically': obj.property

      • 'Dynamically': obj['property']


      By using the 'static' way to access a property, of course, TypeScript will raise an error!



      But with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.



      That's why it will be implicitly marked as any.



      As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny, please refer to his answer for more details about this!



      Hoping this helps ;)






      share|improve this answer





















      • 4





        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
        2 days ago











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

        – Florian
        2 days ago






      • 1





        "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

        – vlaz
        2 days ago











      • @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

        – sjahan
        2 days ago














      28












      28








      28







      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);





      EDIT



      Here is an attempt of an answer for the TypeScript side of the question.



      As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:




      • 'Statically': obj.property

      • 'Dynamically': obj['property']


      By using the 'static' way to access a property, of course, TypeScript will raise an error!



      But with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.



      That's why it will be implicitly marked as any.



      As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny, please refer to his answer for more details about this!



      Hoping this helps ;)






      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);





      EDIT



      Here is an attempt of an answer for the TypeScript side of the question.



      As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:




      • 'Statically': obj.property

      • 'Dynamically': obj['property']


      By using the 'static' way to access a property, of course, TypeScript will raise an error!



      But with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.



      That's why it will be implicitly marked as any.



      As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny, please refer to his answer for more details about this!



      Hoping this helps ;)






      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








      edited 2 days ago

























      answered 2 days ago









      sjahansjahan

      3,2802928




      3,2802928








      • 4





        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
        2 days ago











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

        – Florian
        2 days ago






      • 1





        "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

        – vlaz
        2 days ago











      • @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

        – sjahan
        2 days ago














      • 4





        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
        2 days ago











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

        – Florian
        2 days ago






      • 1





        "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

        – vlaz
        2 days ago











      • @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

        – sjahan
        2 days ago








      4




      4





      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
      2 days 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
      2 days ago













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

      – Florian
      2 days ago





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

      – Florian
      2 days ago




      1




      1





      "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

      – vlaz
      2 days ago





      "Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.

      – vlaz
      2 days ago













      @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

      – sjahan
      2 days ago





      @vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.

      – sjahan
      2 days ago













      15














      It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:



      noImplicitAny enabled



      The reason is that an element access expression returns an object typed as any when the type has no index signature defined (this is an implicit any).



      enter image description here



      So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.






      share|improve this answer






























        15














        It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:



        noImplicitAny enabled



        The reason is that an element access expression returns an object typed as any when the type has no index signature defined (this is an implicit any).



        enter image description here



        So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.






        share|improve this answer




























          15












          15








          15







          It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:



          noImplicitAny enabled



          The reason is that an element access expression returns an object typed as any when the type has no index signature defined (this is an implicit any).



          enter image description here



          So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.






          share|improve this answer















          It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:



          noImplicitAny enabled



          The reason is that an element access expression returns an object typed as any when the type has no index signature defined (this is an implicit any).



          enter image description here



          So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          David SherretDavid Sherret

          51.1k16120123




          51.1k16120123























              4














              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
                2 days ago











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

                – 0xc14m1z
                2 days ago
















              4














              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
                2 days ago











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

                – 0xc14m1z
                2 days ago














              4












              4








              4







              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 2 days ago

























              answered 2 days ago









              0xc14m1z0xc14m1z

              1,534613




              1,534613








              • 1





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

                – Pavlo
                2 days ago











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

                – 0xc14m1z
                2 days ago














              • 1





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

                – Pavlo
                2 days ago











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

                – 0xc14m1z
                2 days ago








              1




              1





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

              – Pavlo
              2 days ago





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

              – Pavlo
              2 days ago













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

              – 0xc14m1z
              2 days ago





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

              – 0xc14m1z
              2 days ago











              1














              The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject) function.



              What happened when you used square braces, someArray.indexOf[someObject], was that the JS runtime first converted your object someObject into a string by calling the function someObject.toString, which most likely returned "[object object]". Then the someArray.indexOf object was queried for the key "[object object]" which wasn't present, returning undefined. As far as Typescript goes, this is completely fine.



              David Sherret pointed out that --noImplicitAny would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.






              share|improve this answer




























                1














                The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject) function.



                What happened when you used square braces, someArray.indexOf[someObject], was that the JS runtime first converted your object someObject into a string by calling the function someObject.toString, which most likely returned "[object object]". Then the someArray.indexOf object was queried for the key "[object object]" which wasn't present, returning undefined. As far as Typescript goes, this is completely fine.



                David Sherret pointed out that --noImplicitAny would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.






                share|improve this answer


























                  1












                  1








                  1







                  The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject) function.



                  What happened when you used square braces, someArray.indexOf[someObject], was that the JS runtime first converted your object someObject into a string by calling the function someObject.toString, which most likely returned "[object object]". Then the someArray.indexOf object was queried for the key "[object object]" which wasn't present, returning undefined. As far as Typescript goes, this is completely fine.



                  David Sherret pointed out that --noImplicitAny would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.






                  share|improve this answer













                  The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject) function.



                  What happened when you used square braces, someArray.indexOf[someObject], was that the JS runtime first converted your object someObject into a string by calling the function someObject.toString, which most likely returned "[object object]". Then the someArray.indexOf object was queried for the key "[object object]" which wasn't present, returning undefined. As far as Typescript goes, this is completely fine.



                  David Sherret pointed out that --noImplicitAny would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 2 days ago









                  Alex SteinbergAlex Steinberg

                  429411




                  429411






























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