Avoid typescript casting inside a switch
2
Consider the following code:
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export interface FooBarAction<T extends FooBarTypes> {
type: T;
data: FooBarTypeMap[T];
}
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction((action as FooBarAction<"FOO">));
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
Now I would like to avoid the casting (action as FooBarAction<"FOO">) as seen in doSomthingBasedOnType, as the very definition if the interface makes this the only possibility inside this switch. Is there something I can change in my code for this to work, or is this simply a bug in TypeScript?
typescript casting typescript-typings typescript-generics
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
add a comment |
2
Consider the following code:
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export interface FooBarAction<T extends FooBarTypes> {
type: T;
data: FooBarTypeMap[T];
}
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction((action as FooBarAction<"FOO">));
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
Now I would like to avoid the casting (action as FooBarAction<"FOO">) as seen in doSomthingBasedOnType, as the very definition if the interface makes this the only possibility inside this switch. Is there something I can change in my code for this to work, or is this simply a bug in TypeScript?
typescript casting typescript-typings typescript-generics
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
add a comment |
2
2
2
Consider the following code:
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export interface FooBarAction<T extends FooBarTypes> {
type: T;
data: FooBarTypeMap[T];
}
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction((action as FooBarAction<"FOO">));
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
Now I would like to avoid the casting (action as FooBarAction<"FOO">) as seen in doSomthingBasedOnType, as the very definition if the interface makes this the only possibility inside this switch. Is there something I can change in my code for this to work, or is this simply a bug in TypeScript?
typescript casting typescript-typings typescript-generics
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
Consider the following code:
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export interface FooBarAction<T extends FooBarTypes> {
type: T;
data: FooBarTypeMap[T];
}
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction((action as FooBarAction<"FOO">));
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
Now I would like to avoid the casting (action as FooBarAction<"FOO">) as seen in doSomthingBasedOnType, as the very definition if the interface makes this the only possibility inside this switch. Is there something I can change in my code for this to work, or is this simply a bug in TypeScript?
typescript casting typescript-typings typescript-generics
typescript casting typescript-typings typescript-generics
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
asked Nov 20 '18 at 11:56
MrMamenMrMamen
508
508
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
You need to transform FooBarAction into a discriminated union. At the moment your version of FooBarAction is not very strict, while type must be one of "FOO" | "BAR" and data must be one of FooBarTypeMap[FooBarTypes] = FooInterface | BarInterface there is no relation between the two. So this could be allowed:
let o : FooBarAction2<FooBarTypes> = {
type: "BAR",
data: {} as FooInterface
}
The discriminated union version would look like this:
export type FooBarAction = {
type: "FOO";
data: FooInterface;
} | {
type: "BAR";
data: BarInterface;
}
const doSomthingBasedOnType = (action: FooBarAction): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
// We use extract to get a specific type from the union
const FooAction = (action: Extract<FooBarAction, { type: "FOO" }>): void => {
//do something with action.data
};
You can also create a union from a union of types using the distributive behavior of conditional types:
interface FooInterface { foo: number}
interface BarInterface { bar: number}
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export type FooBarAction<T extends FooBarTypes> = T extends any ? {
type: T;
data: FooBarTypeMap[T];
}: never;
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })
– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
add a comment |
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1 Answer
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1 Answer
1
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oldest
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2
You need to transform FooBarAction into a discriminated union. At the moment your version of FooBarAction is not very strict, while type must be one of "FOO" | "BAR" and data must be one of FooBarTypeMap[FooBarTypes] = FooInterface | BarInterface there is no relation between the two. So this could be allowed:
let o : FooBarAction2<FooBarTypes> = {
type: "BAR",
data: {} as FooInterface
}
The discriminated union version would look like this:
export type FooBarAction = {
type: "FOO";
data: FooInterface;
} | {
type: "BAR";
data: BarInterface;
}
const doSomthingBasedOnType = (action: FooBarAction): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
// We use extract to get a specific type from the union
const FooAction = (action: Extract<FooBarAction, { type: "FOO" }>): void => {
//do something with action.data
};
You can also create a union from a union of types using the distributive behavior of conditional types:
interface FooInterface { foo: number}
interface BarInterface { bar: number}
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export type FooBarAction<T extends FooBarTypes> = T extends any ? {
type: T;
data: FooBarTypeMap[T];
}: never;
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })
– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
add a comment |
2
You need to transform FooBarAction into a discriminated union. At the moment your version of FooBarAction is not very strict, while type must be one of "FOO" | "BAR" and data must be one of FooBarTypeMap[FooBarTypes] = FooInterface | BarInterface there is no relation between the two. So this could be allowed:
let o : FooBarAction2<FooBarTypes> = {
type: "BAR",
data: {} as FooInterface
}
The discriminated union version would look like this:
export type FooBarAction = {
type: "FOO";
data: FooInterface;
} | {
type: "BAR";
data: BarInterface;
}
const doSomthingBasedOnType = (action: FooBarAction): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
// We use extract to get a specific type from the union
const FooAction = (action: Extract<FooBarAction, { type: "FOO" }>): void => {
//do something with action.data
};
You can also create a union from a union of types using the distributive behavior of conditional types:
interface FooInterface { foo: number}
interface BarInterface { bar: number}
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export type FooBarAction<T extends FooBarTypes> = T extends any ? {
type: T;
data: FooBarTypeMap[T];
}: never;
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })
– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
add a comment |
2
2
2
You need to transform FooBarAction into a discriminated union. At the moment your version of FooBarAction is not very strict, while type must be one of "FOO" | "BAR" and data must be one of FooBarTypeMap[FooBarTypes] = FooInterface | BarInterface there is no relation between the two. So this could be allowed:
let o : FooBarAction2<FooBarTypes> = {
type: "BAR",
data: {} as FooInterface
}
The discriminated union version would look like this:
export type FooBarAction = {
type: "FOO";
data: FooInterface;
} | {
type: "BAR";
data: BarInterface;
}
const doSomthingBasedOnType = (action: FooBarAction): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
// We use extract to get a specific type from the union
const FooAction = (action: Extract<FooBarAction, { type: "FOO" }>): void => {
//do something with action.data
};
You can also create a union from a union of types using the distributive behavior of conditional types:
interface FooInterface { foo: number}
interface BarInterface { bar: number}
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export type FooBarAction<T extends FooBarTypes> = T extends any ? {
type: T;
data: FooBarTypeMap[T];
}: never;
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
You need to transform FooBarAction into a discriminated union. At the moment your version of FooBarAction is not very strict, while type must be one of "FOO" | "BAR" and data must be one of FooBarTypeMap[FooBarTypes] = FooInterface | BarInterface there is no relation between the two. So this could be allowed:
let o : FooBarAction2<FooBarTypes> = {
type: "BAR",
data: {} as FooInterface
}
The discriminated union version would look like this:
export type FooBarAction = {
type: "FOO";
data: FooInterface;
} | {
type: "BAR";
data: BarInterface;
}
const doSomthingBasedOnType = (action: FooBarAction): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
// We use extract to get a specific type from the union
const FooAction = (action: Extract<FooBarAction, { type: "FOO" }>): void => {
//do something with action.data
};
You can also create a union from a union of types using the distributive behavior of conditional types:
interface FooInterface { foo: number}
interface BarInterface { bar: number}
interface FooBarTypeMap {
FOO: FooInterface;
BAR: BarInterface;
}
type FooBarTypes = "FOO" | "BAR";
export type FooBarAction<T extends FooBarTypes> = T extends any ? {
type: T;
data: FooBarTypeMap[T];
}: never;
const doSomthingBasedOnType = (action: FooBarAction<FooBarTypes>): void => {
switch (action.type) {
case "FOO":
FooAction(action);
}
};
const FooAction = (action: FooBarAction<"FOO">): void => {
//do something with action.data
};
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
edited Nov 20 '18 at 12:11
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
answered Nov 20 '18 at 12:04
Titian Cernicova-DragomirTitian Cernicova-Dragomir
65.6k34361
65.6k34361
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })
– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
add a comment |
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })
– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
Seems to be working, but could you explain this part: T extends any ? { type: T; data: FooBarTypeMap[T]; }: never;
– MrMamen
Nov 20 '18 at 13:41
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
@MrMamen This is called "Distributive conditional types" you can read more about it here: typescriptlang.org/docs/handbook/advanced-types.html
– Titian Cernicova-Dragomir
Nov 20 '18 at 13:43
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.
interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })– MrMamen
Nov 22 '18 at 9:31
I've expanded the interfaces to include their own type. Using this I was hoping I could generate a (generic) FooBarAction, but I'm struggeling with typings.
interface FooInterface { foo: number, type: "FOO"} interface BarInterface { bar: number, type: "BAR"} const createFooBarAction = (fooBarData: FooBarTypeMap[FooBarTypes]): FooBarAction<FooBarTypes> => ({ type: fooBarData.type, data: fooBarData })– MrMamen
Nov 22 '18 at 9:31
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
@MrMamen could you please post it as a new question with the full code ? I can't answer right now maybe someone else will and the comments are a really bad place to put a lot of code :)
– Titian Cernicova-Dragomir
Nov 22 '18 at 9:33
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
Good point, created one here: stackoverflow.com/questions/53428841/…
– MrMamen
Nov 22 '18 at 10:28
add a comment |
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