Typescript Generic Union
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1
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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
asked Nov 15 at 18:24
Amol Gupta
648
add a comment |
up vote
1
down vote
favorite
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
asked Nov 15 at 18:24
Amol Gupta
648
“Error On This line” What is the error?
– MTCoster
Nov 15 at 18:28
@MTCoster The error (the important part anyway) isArgument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.
– Titian Cernicova-Dragomir
Nov 15 at 18:34
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 15 at 18:24
Amol Gupta
648
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
typescript typescript-typings typescript-generics
asked Nov 15 at 18:24
Amol Gupta
648
asked Nov 15 at 18:24
Amol Gupta
648
edited Nov 15 at 18:34
asked Nov 15 at 18:24
Amol Gupta
648
asked Nov 15 at 18:24
Amol Gupta
648
asked Nov 15 at 18:24
Amol Gupta
648
648
“Error On This line” What is the error?
– MTCoster
Nov 15 at 18:28
@MTCoster The error (the important part anyway) isArgument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.
– Titian Cernicova-Dragomir
Nov 15 at 18:34
add a comment |
“Error On This line” What is the error?
– MTCoster
Nov 15 at 18:28
@MTCoster The error (the important part anyway) isArgument of type 'someGeneric<string> | someGeneric<number>' is not assignable to parameter of type 'someGeneric<string>'.
– Titian Cernicova-Dragomir
Nov 15 at 18:34
“Error On This line” What is the error?
– MTCoster
Nov 15 at 18:28
“Error On This line” What is the error?
– MTCoster
Nov 15 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 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 at 18:34
add a comment |
1 Answer
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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 => {
someFunction(generic); // retruns string | number
});
Playground link
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 at 18:41
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
up vote
2
down vote
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 => {
someFunction(generic); // retruns string | number
});
Playground link
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 at 18:41
add a comment |
up vote
2
down vote
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 => {
someFunction(generic); // retruns string | number
});
Playground link
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 at 18:41
add a comment |
up vote
2
down vote
up vote
2
down vote
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 => {
someFunction(generic); // retruns string | number
});
Playground link
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
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 => {
someFunction(generic); // retruns string | number
});
Playground link
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
answered Nov 15 at 18:32
Titian Cernicova-Dragomir
55.2k33351
55.2k33351
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 at 18:41
add a comment |
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 at 18:41
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 at 18:40
Ahh! Beautiful thanks mate!!! God Bless.
– Amol Gupta
Nov 15 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 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 at 18:41
add a comment |
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“Error On This line” What is the error?
– MTCoster
Nov 15 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 at 18:34