Generic type of extended interface not inferred
In the following example Typescript can infer the type of T in method foo from the parameter passed to it in bar, but it doesn't infer the type of R, which it feels like it should - given that it knows the type of T and also that T extends I<R>
interface I<T> {
}
class A implements I<string> {
}
function foo<T extends I<R>, R>(bar: T): R {
return;
}
foo(new A());
Is there another way to do this?
typescript generics types type-inference
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
add a comment |
In the following example Typescript can infer the type of T in method foo from the parameter passed to it in bar, but it doesn't infer the type of R, which it feels like it should - given that it knows the type of T and also that T extends I<R>
interface I<T> {
}
class A implements I<string> {
}
function foo<T extends I<R>, R>(bar: T): R {
return;
}
foo(new A());
Is there another way to do this?
typescript generics types type-inference
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
add a comment |
In the following example Typescript can infer the type of T in method foo from the parameter passed to it in bar, but it doesn't infer the type of R, which it feels like it should - given that it knows the type of T and also that T extends I<R>
interface I<T> {
}
class A implements I<string> {
}
function foo<T extends I<R>, R>(bar: T): R {
return;
}
foo(new A());
Is there another way to do this?
typescript generics types type-inference
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
In the following example Typescript can infer the type of T in method foo from the parameter passed to it in bar, but it doesn't infer the type of R, which it feels like it should - given that it knows the type of T and also that T extends I<R>
interface I<T> {
}
class A implements I<string> {
}
function foo<T extends I<R>, R>(bar: T): R {
return;
}
foo(new A());
Is there another way to do this?
typescript generics types type-inference
typescript generics types type-inference
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
asked Nov 23 '18 at 14:01
IBOED2IBOED2
54116
54116
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The first problem is that your interface is empty, typescript uses structural typing so if your generic interface will not use it's type parameter it won't matter much that it has it. For example this works:
interface I<T> { }
declare let foo: I<string>
declare let bar: I<number>
// Same structure ({}), assignment works
foo = bar
bar = foo
Event if we add a field, Typescript will still not infer the R type parameter, it just doesn't try to extract it from T. Your best option is to use a conditional type and extract the generic type parameter where you need it:
interface I<T> {
value :T
}
class A implements I<string> {
value! :string
}
type ExtractFromI<T extends I<unknown>> = T extends I<infer U> ? U : never;
function foo<T extends I<unknown>>(bar: T): ExtractFromI<T>{
return bar.value as ExtractFromI<T>; // Generic type with unresolved type parameters, we need a type assertion to convince the compiler this is ok
}
var r = foo(new A()); // string
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
That's super cool thanks, follow up question, is it possible to make thisExtractFrommapped type generic? So that it can extract from any interface rather than specifically I?
– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The first problem is that your interface is empty, typescript uses structural typing so if your generic interface will not use it's type parameter it won't matter much that it has it. For example this works:
interface I<T> { }
declare let foo: I<string>
declare let bar: I<number>
// Same structure ({}), assignment works
foo = bar
bar = foo
Event if we add a field, Typescript will still not infer the R type parameter, it just doesn't try to extract it from T. Your best option is to use a conditional type and extract the generic type parameter where you need it:
interface I<T> {
value :T
}
class A implements I<string> {
value! :string
}
type ExtractFromI<T extends I<unknown>> = T extends I<infer U> ? U : never;
function foo<T extends I<unknown>>(bar: T): ExtractFromI<T>{
return bar.value as ExtractFromI<T>; // Generic type with unresolved type parameters, we need a type assertion to convince the compiler this is ok
}
var r = foo(new A()); // string
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
That's super cool thanks, follow up question, is it possible to make thisExtractFrommapped type generic? So that it can extract from any interface rather than specifically I?
– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
add a comment |
The first problem is that your interface is empty, typescript uses structural typing so if your generic interface will not use it's type parameter it won't matter much that it has it. For example this works:
interface I<T> { }
declare let foo: I<string>
declare let bar: I<number>
// Same structure ({}), assignment works
foo = bar
bar = foo
Event if we add a field, Typescript will still not infer the R type parameter, it just doesn't try to extract it from T. Your best option is to use a conditional type and extract the generic type parameter where you need it:
interface I<T> {
value :T
}
class A implements I<string> {
value! :string
}
type ExtractFromI<T extends I<unknown>> = T extends I<infer U> ? U : never;
function foo<T extends I<unknown>>(bar: T): ExtractFromI<T>{
return bar.value as ExtractFromI<T>; // Generic type with unresolved type parameters, we need a type assertion to convince the compiler this is ok
}
var r = foo(new A()); // string
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
That's super cool thanks, follow up question, is it possible to make thisExtractFrommapped type generic? So that it can extract from any interface rather than specifically I?
– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
add a comment |
The first problem is that your interface is empty, typescript uses structural typing so if your generic interface will not use it's type parameter it won't matter much that it has it. For example this works:
interface I<T> { }
declare let foo: I<string>
declare let bar: I<number>
// Same structure ({}), assignment works
foo = bar
bar = foo
Event if we add a field, Typescript will still not infer the R type parameter, it just doesn't try to extract it from T. Your best option is to use a conditional type and extract the generic type parameter where you need it:
interface I<T> {
value :T
}
class A implements I<string> {
value! :string
}
type ExtractFromI<T extends I<unknown>> = T extends I<infer U> ? U : never;
function foo<T extends I<unknown>>(bar: T): ExtractFromI<T>{
return bar.value as ExtractFromI<T>; // Generic type with unresolved type parameters, we need a type assertion to convince the compiler this is ok
}
var r = foo(new A()); // string
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
The first problem is that your interface is empty, typescript uses structural typing so if your generic interface will not use it's type parameter it won't matter much that it has it. For example this works:
interface I<T> { }
declare let foo: I<string>
declare let bar: I<number>
// Same structure ({}), assignment works
foo = bar
bar = foo
Event if we add a field, Typescript will still not infer the R type parameter, it just doesn't try to extract it from T. Your best option is to use a conditional type and extract the generic type parameter where you need it:
interface I<T> {
value :T
}
class A implements I<string> {
value! :string
}
type ExtractFromI<T extends I<unknown>> = T extends I<infer U> ? U : never;
function foo<T extends I<unknown>>(bar: T): ExtractFromI<T>{
return bar.value as ExtractFromI<T>; // Generic type with unresolved type parameters, we need a type assertion to convince the compiler this is ok
}
var r = foo(new A()); // string
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
answered Nov 23 '18 at 14:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
59.2k33553
59.2k33553
That's super cool thanks, follow up question, is it possible to make thisExtractFrommapped type generic? So that it can extract from any interface rather than specifically I?
– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
add a comment |
That's super cool thanks, follow up question, is it possible to make thisExtractFrommapped type generic? So that it can extract from any interface rather than specifically I?
– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
That's super cool thanks, follow up question, is it possible to make this
ExtractFrom mapped type generic? So that it can extract from any interface rather than specifically I?– IBOED2
Nov 23 '18 at 14:52
That's super cool thanks, follow up question, is it possible to make this
ExtractFrom mapped type generic? So that it can extract from any interface rather than specifically I?– IBOED2
Nov 23 '18 at 14:52
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
@IBOED2 unfortunately no, you will need a dedicated conditional type tailored for each generic type you want to extract parameters from.
– Titian Cernicova-Dragomir
Nov 23 '18 at 14:53
add a comment |
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