Typescript type inference from type parameters
I don't understand why Typescript can't properly infer the types in the following case that involves inference from type parameters. (This question is similar to TypeScript type inference issue but is a somewhat different case. The answer may be the same, though, that I'm just out of luck!)
// A base class for a dialog taking parameter of type P
// and returning result of type R.
class BaseDialog<P, R> { p: P; r: R; }
class ValueDialog extends BaseDialog<string, number> {}
// A function that shows the dialog
show<T extends BaseDialog<P, R>, P, R>(dlg: Type<T>, param: P): Promise<R> {}
Note: To simplify the method signature, I'm using Angular's Type:
export interface Type<T> extends Function {
new (...args: any): T;
}
Now, when I call the method show, as follows, the R type is not correctly inferred:
show(ValueDialog, "name").then(r => console.log(r));
The compiler infers:
T = ValueDialog
P = string
R = {}
Since T is correctly inferred, you'd think the compiler could infer P and R from ValueDialog's definition, but it doesn't.
I can fix this, of course, by manually specifying the types, but that's pretty ugly. I can also fix it by making P and R the same, but that isn't the functionality I want.
How can I define show() so that it correctly infers R?
typescript type-inference
asked Nov 23 '18 at 15:06
RDGRDG
386
add a comment |
I don't understand why Typescript can't properly infer the types in the following case that involves inference from type parameters. (This question is similar to TypeScript type inference issue but is a somewhat different case. The answer may be the same, though, that I'm just out of luck!)
// A base class for a dialog taking parameter of type P
// and returning result of type R.
class BaseDialog<P, R> { p: P; r: R; }
class ValueDialog extends BaseDialog<string, number> {}
// A function that shows the dialog
show<T extends BaseDialog<P, R>, P, R>(dlg: Type<T>, param: P): Promise<R> {}
Note: To simplify the method signature, I'm using Angular's Type:
export interface Type<T> extends Function {
new (...args: any): T;
}
Now, when I call the method show, as follows, the R type is not correctly inferred:
show(ValueDialog, "name").then(r => console.log(r));
The compiler infers:
T = ValueDialog
P = string
R = {}
Since T is correctly inferred, you'd think the compiler could infer P and R from ValueDialog's definition, but it doesn't.
I can fix this, of course, by manually specifying the types, but that's pretty ugly. I can also fix it by making P and R the same, but that isn't the functionality I want.
How can I define show() so that it correctly infers R?
typescript type-inference
asked Nov 23 '18 at 15:06
RDGRDG
386
This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07
add a comment |
I don't understand why Typescript can't properly infer the types in the following case that involves inference from type parameters. (This question is similar to TypeScript type inference issue but is a somewhat different case. The answer may be the same, though, that I'm just out of luck!)
// A base class for a dialog taking parameter of type P
// and returning result of type R.
class BaseDialog<P, R> { p: P; r: R; }
class ValueDialog extends BaseDialog<string, number> {}
// A function that shows the dialog
show<T extends BaseDialog<P, R>, P, R>(dlg: Type<T>, param: P): Promise<R> {}
Note: To simplify the method signature, I'm using Angular's Type:
export interface Type<T> extends Function {
new (...args: any): T;
}
Now, when I call the method show, as follows, the R type is not correctly inferred:
show(ValueDialog, "name").then(r => console.log(r));
The compiler infers:
T = ValueDialog
P = string
R = {}
Since T is correctly inferred, you'd think the compiler could infer P and R from ValueDialog's definition, but it doesn't.
I can fix this, of course, by manually specifying the types, but that's pretty ugly. I can also fix it by making P and R the same, but that isn't the functionality I want.
How can I define show() so that it correctly infers R?
typescript type-inference
asked Nov 23 '18 at 15:06
RDGRDG
386
I don't understand why Typescript can't properly infer the types in the following case that involves inference from type parameters. (This question is similar to TypeScript type inference issue but is a somewhat different case. The answer may be the same, though, that I'm just out of luck!)
// A base class for a dialog taking parameter of type P
// and returning result of type R.
class BaseDialog<P, R> { p: P; r: R; }
class ValueDialog extends BaseDialog<string, number> {}
// A function that shows the dialog
show<T extends BaseDialog<P, R>, P, R>(dlg: Type<T>, param: P): Promise<R> {}
Note: To simplify the method signature, I'm using Angular's Type:
export interface Type<T> extends Function {
new (...args: any): T;
}
Now, when I call the method show, as follows, the R type is not correctly inferred:
show(ValueDialog, "name").then(r => console.log(r));
The compiler infers:
T = ValueDialog
P = string
R = {}
Since T is correctly inferred, you'd think the compiler could infer P and R from ValueDialog's definition, but it doesn't.
I can fix this, of course, by manually specifying the types, but that's pretty ugly. I can also fix it by making P and R the same, but that isn't the functionality I want.
How can I define show() so that it correctly infers R?
typescript type-inference
typescript type-inference
asked Nov 23 '18 at 15:06
RDGRDG
386
asked Nov 23 '18 at 15:06
RDGRDG
386
edited Nov 23 '18 at 15:35
RDG
asked Nov 23 '18 at 15:06
RDGRDG
386
asked Nov 23 '18 at 15:06
RDGRDG
386
asked Nov 23 '18 at 15:06
RDGRDG
386
386
This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07
add a comment |
This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07
This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07
This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07
add a comment |
1 Answer
1
active
oldest
votes
You ca use a conditional type to extract type R parameter from the base type. You will need to use the R type in some way for this to work :
export interface Type<T> extends Function {
new (...args: any): T;
}
class BaseDialog<P, R> {
value: R //we need to use R in some way for the parameter to make a difference
}
class ValueDialog extends BaseDialog<string, number> {}
type ExtractResult<T extends BaseDialog<any, any>> = T extends BaseDialog<any, infer R> ? R : never;
// A function that shows the dialog
declare function show<T extends BaseDialog<P, any>, P>(dlg: Type<T>, param: P): Promise<ExtractResult<T>>;
show(ValueDialog, "name").then(r => console.log(r)); // r is now string
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You ca use a conditional type to extract type R parameter from the base type. You will need to use the R type in some way for this to work :
export interface Type<T> extends Function {
new (...args: any): T;
}
class BaseDialog<P, R> {
value: R //we need to use R in some way for the parameter to make a difference
}
class ValueDialog extends BaseDialog<string, number> {}
type ExtractResult<T extends BaseDialog<any, any>> = T extends BaseDialog<any, infer R> ? R : never;
// A function that shows the dialog
declare function show<T extends BaseDialog<P, any>, P>(dlg: Type<T>, param: P): Promise<ExtractResult<T>>;
show(ValueDialog, "name").then(r => console.log(r)); // r is now string
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
add a comment |
You ca use a conditional type to extract type R parameter from the base type. You will need to use the R type in some way for this to work :
export interface Type<T> extends Function {
new (...args: any): T;
}
class BaseDialog<P, R> {
value: R //we need to use R in some way for the parameter to make a difference
}
class ValueDialog extends BaseDialog<string, number> {}
type ExtractResult<T extends BaseDialog<any, any>> = T extends BaseDialog<any, infer R> ? R : never;
// A function that shows the dialog
declare function show<T extends BaseDialog<P, any>, P>(dlg: Type<T>, param: P): Promise<ExtractResult<T>>;
show(ValueDialog, "name").then(r => console.log(r)); // r is now string
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
add a comment |
You ca use a conditional type to extract type R parameter from the base type. You will need to use the R type in some way for this to work :
export interface Type<T> extends Function {
new (...args: any): T;
}
class BaseDialog<P, R> {
value: R //we need to use R in some way for the parameter to make a difference
}
class ValueDialog extends BaseDialog<string, number> {}
type ExtractResult<T extends BaseDialog<any, any>> = T extends BaseDialog<any, infer R> ? R : never;
// A function that shows the dialog
declare function show<T extends BaseDialog<P, any>, P>(dlg: Type<T>, param: P): Promise<ExtractResult<T>>;
show(ValueDialog, "name").then(r => console.log(r)); // r is now string
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
You ca use a conditional type to extract type R parameter from the base type. You will need to use the R type in some way for this to work :
export interface Type<T> extends Function {
new (...args: any): T;
}
class BaseDialog<P, R> {
value: R //we need to use R in some way for the parameter to make a difference
}
class ValueDialog extends BaseDialog<string, number> {}
type ExtractResult<T extends BaseDialog<any, any>> = T extends BaseDialog<any, infer R> ? R : never;
// A function that shows the dialog
declare function show<T extends BaseDialog<P, any>, P>(dlg: Type<T>, param: P): Promise<ExtractResult<T>>;
show(ValueDialog, "name").then(r => console.log(r)); // r is now string
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
answered Nov 23 '18 at 15:12
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
65.6k34361
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
add a comment |
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
1
1
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
This works! I can (mostly) follow the ExtractResult type, but I could never have come up with this on my own! I'm from the other link that this is the only way to infer types from a type parameter. Thank you for the help!
– RDG
Nov 24 '18 at 19:54
add a comment |
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This might help: stackoverflow.com/questions/53448100/…
– Titian Cernicova-Dragomir
Nov 23 '18 at 15:07