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?










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?










    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?










      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






      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
          1






          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























            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%2f53996797%2ftypescript-conditional-type-array-of-union-type-distribution%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            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
































                    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%2f53996797%2ftypescript-conditional-type-array-of-union-type-distribution%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

                    Monofisismo

                    Angular Downloading a file using contenturl with Basic Authentication

                    Olmecas