TypeScript: Conditional type array of union type distribution












2















I have a conditional type that uses a generic type T to determine an Array<T> type. As a contrived example:



type X<T> = T extends string ? Array<T> : never;


The issue I am having is when I provide a union type, it is being distributed as a union of 2 array types instead of an array of my union type.



// compiler complains because it expects Array<'one'> | Array<'two'>

const y: X<'one' | 'two'> = ['one', 'two'];


Is there a way to type this such that my conditional type produces an Array<'one' | 'two'> if the condition is met?






typescript discriminated-union







share|improve this question



























    2















    I have a conditional type that uses a generic type T to determine an Array<T> type. As a contrived example:



    type X<T> = T extends string ? Array<T> : never;


    The issue I am having is when I provide a union type, it is being distributed as a union of 2 array types instead of an array of my union type.



    // compiler complains because it expects Array<'one'> | Array<'two'>
    
const y: X<'one' | 'two'> = ['one', 'two'];


    Is there a way to type this such that my conditional type produces an Array<'one' | 'two'> if the condition is met?






    typescript discriminated-union







    share|improve this question

























      2












      2








      2








      I have a conditional type that uses a generic type T to determine an Array<T> type. As a contrived example:



      type X<T> = T extends string ? Array<T> : never;


      The issue I am having is when I provide a union type, it is being distributed as a union of 2 array types instead of an array of my union type.



      // compiler complains because it expects Array<'one'> | Array<'two'>
      
const y: X<'one' | 'two'> = ['one', 'two'];


      Is there a way to type this such that my conditional type produces an Array<'one' | 'two'> if the condition is met?






      typescript discriminated-union







      share|improve this question














      I have a conditional type that uses a generic type T to determine an Array<T> type. As a contrived example:



      type X<T> = T extends string ? Array<T> : never;


      The issue I am having is when I provide a union type, it is being distributed as a union of 2 array types instead of an array of my union type.



      // compiler complains because it expects Array<'one'> | Array<'two'>
      
const y: X<'one' | 'two'> = ['one', 'two'];


      Is there a way to type this such that my conditional type produces an Array<'one' | 'two'> if the condition is met?






      typescript discriminated-union




      typescript discriminated-union






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 1 at 15:47









      binglesbingles

      5,41644858




      5,41644858
























          1 Answer
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          You have run into the distributive behavior of conditional types where a conditional type is distributed over a naked type parameter containing a union. This behavior is very useful in some scenarios but can be a bit surprising at first.



          The simples option to disable this behavior is to put the type parameter in a tuple:



          type X<T> = [T] extends [string] ? Array<T> : never;
          

          
// ok y is Array<'one' | 'two'>
          
const y: X<'one' | 'two'> = ['one', 'two'];


          You can read more about this behavior here and here






          share|improve this answer























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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3














            You have run into the distributive behavior of conditional types where a conditional type is distributed over a naked type parameter containing a union. This behavior is very useful in some scenarios but can be a bit surprising at first.



            The simples option to disable this behavior is to put the type parameter in a tuple:



            type X<T> = [T] extends [string] ? Array<T> : never;
            

            
// ok y is Array<'one' | 'two'>
            
const y: X<'one' | 'two'> = ['one', 'two'];


            You can read more about this behavior here and here






            share|improve this answer




























              3














              You have run into the distributive behavior of conditional types where a conditional type is distributed over a naked type parameter containing a union. This behavior is very useful in some scenarios but can be a bit surprising at first.



              The simples option to disable this behavior is to put the type parameter in a tuple:



              type X<T> = [T] extends [string] ? Array<T> : never;
              

              
// ok y is Array<'one' | 'two'>
              
const y: X<'one' | 'two'> = ['one', 'two'];


              You can read more about this behavior here and here






              share|improve this answer


























                3












                3








                3







                You have run into the distributive behavior of conditional types where a conditional type is distributed over a naked type parameter containing a union. This behavior is very useful in some scenarios but can be a bit surprising at first.



                The simples option to disable this behavior is to put the type parameter in a tuple:



                type X<T> = [T] extends [string] ? Array<T> : never;
                

                
// ok y is Array<'one' | 'two'>
                
const y: X<'one' | 'two'> = ['one', 'two'];


                You can read more about this behavior here and here






                share|improve this answer













                You have run into the distributive behavior of conditional types where a conditional type is distributed over a naked type parameter containing a union. This behavior is very useful in some scenarios but can be a bit surprising at first.



                The simples option to disable this behavior is to put the type parameter in a tuple:



                type X<T> = [T] extends [string] ? Array<T> : never;
                

                
// ok y is Array<'one' | 'two'>
                
const y: X<'one' | 'two'> = ['one', 'two'];


                You can read more about this behavior here and here







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Jan 1 at 15:51









                Titian Cernicova-DragomirTitian Cernicova-Dragomir

                67.4k34563




                67.4k34563
































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