Typescript Generic Union










1















So I have Array of generic object and want to iterate over the but typescript wont allow me. Here is some sample code. Any Suggestions of how this can be solved.



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type someFunction = <T>(generic: someGeneric<T>) => T;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let genericArray = [stringGeneric, numberGeneric];
genericArray.forEach(generic =>
someFunction(generic); // Error On This line.
);


You can just copy-paste the code to this link. I cant seem to share the code.










share|improve this question
























  • Error On This line” What is the error?

    – MTCoster
    Nov 15 '18 at 18:28











  • @MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:34















1















So I have Array of generic object and want to iterate over the but typescript wont allow me. Here is some sample code. Any Suggestions of how this can be solved.



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type someFunction = <T>(generic: someGeneric<T>) => T;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let genericArray = [stringGeneric, numberGeneric];
genericArray.forEach(generic =>
someFunction(generic); // Error On This line.
);


You can just copy-paste the code to this link. I cant seem to share the code.










share|improve this question
























  • Error On This line” What is the error?

    – MTCoster
    Nov 15 '18 at 18:28











  • @MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:34













1












1








1








So I have Array of generic object and want to iterate over the but typescript wont allow me. Here is some sample code. Any Suggestions of how this can be solved.



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type someFunction = <T>(generic: someGeneric<T>) => T;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let genericArray = [stringGeneric, numberGeneric];
genericArray.forEach(generic =>
someFunction(generic); // Error On This line.
);


You can just copy-paste the code to this link. I cant seem to share the code.










share|improve this question
















So I have Array of generic object and want to iterate over the but typescript wont allow me. Here is some sample code. Any Suggestions of how this can be solved.



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type someFunction = <T>(generic: someGeneric<T>) => T;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let genericArray = [stringGeneric, numberGeneric];
genericArray.forEach(generic =>
someFunction(generic); // Error On This line.
);


You can just copy-paste the code to this link. I cant seem to share the code.







typescript typescript-typings typescript-generics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 15 '18 at 18:34







Amol Gupta

















asked Nov 15 '18 at 18:24









Amol GuptaAmol Gupta

1038




1038












  • Error On This line” What is the error?

    – MTCoster
    Nov 15 '18 at 18:28











  • @MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:34

















  • Error On This line” What is the error?

    – MTCoster
    Nov 15 '18 at 18:28











  • @MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:34
















Error On This line” What is the error?

– MTCoster
Nov 15 '18 at 18:28





Error On This line” What is the error?

– MTCoster
Nov 15 '18 at 18:28













@MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

– Titian Cernicova-Dragomir
Nov 15 '18 at 18:34





@MTCoster The error (the important part anyway) is Argument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.

– Titian Cernicova-Dragomir
Nov 15 '18 at 18:34












1 Answer
1






active

oldest

votes


















2














The problem is that the function accepts a parameter of type someGeneric<T> if we try to pass in a parameter of type someGeneric<number> | someGeneric<string> typescript will not try to infer T from this it will just say the union is not compatible with the type someGeneric<T>



We can change the definition of the function so that the type parameter extends someGeneric<any>. This constraint will be compatible with the union. We can then use a conditional type to extract the item type from the T using a conditional type. Since conditional types distribute over unions the result of the extraction will be a union of the generic parameters to someGeneric<T>



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type extractItemFromSomeGeneric<T extends someGeneric<any>> = T extends someGeneric<infer U> ? U : never;
type someFunction = <T extends someGeneric<any>>(generic: T) => extractItemFromSomeGeneric<T>;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let someGeneric = [stringGeneric, numberGeneric];
someGeneric.forEach(generic => number
);


Playground link






share|improve this answer























  • Ahh! Beautiful thanks mate!!! God Bless.

    – Amol Gupta
    Nov 15 '18 at 18:40











  • @AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:41










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The problem is that the function accepts a parameter of type someGeneric<T> if we try to pass in a parameter of type someGeneric<number> | someGeneric<string> typescript will not try to infer T from this it will just say the union is not compatible with the type someGeneric<T>



We can change the definition of the function so that the type parameter extends someGeneric<any>. This constraint will be compatible with the union. We can then use a conditional type to extract the item type from the T using a conditional type. Since conditional types distribute over unions the result of the extraction will be a union of the generic parameters to someGeneric<T>



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type extractItemFromSomeGeneric<T extends someGeneric<any>> = T extends someGeneric<infer U> ? U : never;
type someFunction = <T extends someGeneric<any>>(generic: T) => extractItemFromSomeGeneric<T>;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let someGeneric = [stringGeneric, numberGeneric];
someGeneric.forEach(generic => number
);


Playground link






share|improve this answer























  • Ahh! Beautiful thanks mate!!! God Bless.

    – Amol Gupta
    Nov 15 '18 at 18:40











  • @AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:41















2














The problem is that the function accepts a parameter of type someGeneric<T> if we try to pass in a parameter of type someGeneric<number> | someGeneric<string> typescript will not try to infer T from this it will just say the union is not compatible with the type someGeneric<T>



We can change the definition of the function so that the type parameter extends someGeneric<any>. This constraint will be compatible with the union. We can then use a conditional type to extract the item type from the T using a conditional type. Since conditional types distribute over unions the result of the extraction will be a union of the generic parameters to someGeneric<T>



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type extractItemFromSomeGeneric<T extends someGeneric<any>> = T extends someGeneric<infer U> ? U : never;
type someFunction = <T extends someGeneric<any>>(generic: T) => extractItemFromSomeGeneric<T>;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let someGeneric = [stringGeneric, numberGeneric];
someGeneric.forEach(generic => number
);


Playground link






share|improve this answer























  • Ahh! Beautiful thanks mate!!! God Bless.

    – Amol Gupta
    Nov 15 '18 at 18:40











  • @AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:41













2












2








2







The problem is that the function accepts a parameter of type someGeneric<T> if we try to pass in a parameter of type someGeneric<number> | someGeneric<string> typescript will not try to infer T from this it will just say the union is not compatible with the type someGeneric<T>



We can change the definition of the function so that the type parameter extends someGeneric<any>. This constraint will be compatible with the union. We can then use a conditional type to extract the item type from the T using a conditional type. Since conditional types distribute over unions the result of the extraction will be a union of the generic parameters to someGeneric<T>



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type extractItemFromSomeGeneric<T extends someGeneric<any>> = T extends someGeneric<infer U> ? U : never;
type someFunction = <T extends someGeneric<any>>(generic: T) => extractItemFromSomeGeneric<T>;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let someGeneric = [stringGeneric, numberGeneric];
someGeneric.forEach(generic => number
);


Playground link






share|improve this answer













The problem is that the function accepts a parameter of type someGeneric<T> if we try to pass in a parameter of type someGeneric<number> | someGeneric<string> typescript will not try to infer T from this it will just say the union is not compatible with the type someGeneric<T>



We can change the definition of the function so that the type parameter extends someGeneric<any>. This constraint will be compatible with the union. We can then use a conditional type to extract the item type from the T using a conditional type. Since conditional types distribute over unions the result of the extraction will be a union of the generic parameters to someGeneric<T>



type someGeneric<T> = item: T ;

type stringGeneric = someGeneric<string>;

type numberGeneric = someGeneric<number>;

type extractItemFromSomeGeneric<T extends someGeneric<any>> = T extends someGeneric<infer U> ? U : never;
type someFunction = <T extends someGeneric<any>>(generic: T) => extractItemFromSomeGeneric<T>;

const someFunction: someFunction = (generic) => generic.item;

const stringGeneric: stringGeneric = item: 'some String' ,
numberGeneric: numberGeneric = item: 12 ;

let someGeneric = [stringGeneric, numberGeneric];
someGeneric.forEach(generic => number
);


Playground link







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 15 '18 at 18:32









Titian Cernicova-DragomirTitian Cernicova-Dragomir

71.6k35068




71.6k35068












  • Ahh! Beautiful thanks mate!!! God Bless.

    – Amol Gupta
    Nov 15 '18 at 18:40











  • @AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:41

















  • Ahh! Beautiful thanks mate!!! God Bless.

    – Amol Gupta
    Nov 15 '18 at 18:40











  • @AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

    – Titian Cernicova-Dragomir
    Nov 15 '18 at 18:41
















Ahh! Beautiful thanks mate!!! God Bless.

– Amol Gupta
Nov 15 '18 at 18:40





Ahh! Beautiful thanks mate!!! God Bless.

– Amol Gupta
Nov 15 '18 at 18:40













@AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

– Titian Cernicova-Dragomir
Nov 15 '18 at 18:41





@AmolGupta don't forget to mark as answered, it would really help me to get to 200 for today :D

– Titian Cernicova-Dragomir
Nov 15 '18 at 18:41



















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