Mixing
Type definition for a jagged array of type T and its array prototype extension
2
I'm trying to write a Typescript definition file for a JavaScript function I've attached to Array.prototype.
Array.js
/**
* Flattens the array recursively.
*
* @example
* [1, [2, 3]].flat() // => [1, 2, 3]
*
* @example
* [[1, [2, 3], [4, [5]]], 6].flat() // => [1, 2, 3, 4, 5, 6]
*/
Array.prototype.flat = function () {
return this.reduce((arr, val) => Array.isArray(val) ? arr.concat(val.flat()) : arr.concat(val), );
}
flat() works on an Array<T|S> where S is Array<T|S> and returns Array<T>. That is, it has a recursive definition, and all arrays fit the definition, as [1, 2, 3].flat() will simply return a copy of the original array.
I'm new to TypeScript, but my understanding is that in order to get the benefits of a TypeScript definition file (namely IntelliSense), the method definition must be within interface Array<T>. If this is the case, is there a way to place a constraint on T for a specialized version of Array<T>?
If not, how can I define an interface that will be picked up for every array AND will recognize when the array fits the recursive definition?
javascript typescript typescript-generics
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
add a comment |
2
I'm trying to write a Typescript definition file for a JavaScript function I've attached to Array.prototype.
Array.js
/**
* Flattens the array recursively.
*
* @example
* [1, [2, 3]].flat() // => [1, 2, 3]
*
* @example
* [[1, [2, 3], [4, [5]]], 6].flat() // => [1, 2, 3, 4, 5, 6]
*/
Array.prototype.flat = function () {
return this.reduce((arr, val) => Array.isArray(val) ? arr.concat(val.flat()) : arr.concat(val), );
}
flat() works on an Array<T|S> where S is Array<T|S> and returns Array<T>. That is, it has a recursive definition, and all arrays fit the definition, as [1, 2, 3].flat() will simply return a copy of the original array.
I'm new to TypeScript, but my understanding is that in order to get the benefits of a TypeScript definition file (namely IntelliSense), the method definition must be within interface Array<T>. If this is the case, is there a way to place a constraint on T for a specialized version of Array<T>?
If not, how can I define an interface that will be picked up for every array AND will recognize when the array fits the recursive definition?
javascript typescript typescript-generics
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
add a comment |
2
2
2
I'm trying to write a Typescript definition file for a JavaScript function I've attached to Array.prototype.
Array.js
/**
* Flattens the array recursively.
*
* @example
* [1, [2, 3]].flat() // => [1, 2, 3]
*
* @example
* [[1, [2, 3], [4, [5]]], 6].flat() // => [1, 2, 3, 4, 5, 6]
*/
Array.prototype.flat = function () {
return this.reduce((arr, val) => Array.isArray(val) ? arr.concat(val.flat()) : arr.concat(val), );
}
flat() works on an Array<T|S> where S is Array<T|S> and returns Array<T>. That is, it has a recursive definition, and all arrays fit the definition, as [1, 2, 3].flat() will simply return a copy of the original array.
I'm new to TypeScript, but my understanding is that in order to get the benefits of a TypeScript definition file (namely IntelliSense), the method definition must be within interface Array<T>. If this is the case, is there a way to place a constraint on T for a specialized version of Array<T>?
If not, how can I define an interface that will be picked up for every array AND will recognize when the array fits the recursive definition?
javascript typescript typescript-generics
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
I'm trying to write a Typescript definition file for a JavaScript function I've attached to Array.prototype.
Array.js
/**
* Flattens the array recursively.
*
* @example
* [1, [2, 3]].flat() // => [1, 2, 3]
*
* @example
* [[1, [2, 3], [4, [5]]], 6].flat() // => [1, 2, 3, 4, 5, 6]
*/
Array.prototype.flat = function () {
return this.reduce((arr, val) => Array.isArray(val) ? arr.concat(val.flat()) : arr.concat(val), );
}
flat() works on an Array<T|S> where S is Array<T|S> and returns Array<T>. That is, it has a recursive definition, and all arrays fit the definition, as [1, 2, 3].flat() will simply return a copy of the original array.
I'm new to TypeScript, but my understanding is that in order to get the benefits of a TypeScript definition file (namely IntelliSense), the method definition must be within interface Array<T>. If this is the case, is there a way to place a constraint on T for a specialized version of Array<T>?
If not, how can I define an interface that will be picked up for every array AND will recognize when the array fits the recursive definition?
javascript typescript typescript-generics
javascript typescript typescript-generics
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
edited Nov 20 '18 at 6:23
Shaun Luttin
58.1k32223285
58.1k32223285
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
asked Nov 20 '18 at 2:55
dx_over_dtdx_over_dt
1,60342346
1,60342346
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
This is not a complete answer, but it does get us most of the way. Here it is in the TypeScript Playground.
type JaggedArrayItem<T> = T | JaggedArray<T>;
interface JaggedArray<T> extends Array<JaggedArrayItem<T>> { }
type FlatArray<T> = T extends JaggedArrayItem<infer U> ? U : T;
interface Array<T> {
flat(this: JaggedArray<T>): FlatArray<T>;
}
Array.prototype.flat = function () {
return this.reduce(
(arr, val) => Array.isArray(val)
? arr.concat(val.flat())
: arr.concat(val),
);
};
// Test
const myRecursiveArray: JaggedArray<number> = [
10,
[9],
[
[8],
[
[7],
]
],
];
const flattened: number = myRecursiveArray.flat();
const flattenedToo: (string | number) = [1, 'two'].flat();
See also
- https://github.com/Microsoft/TypeScript/issues/3496#issuecomment-128553540
- https://github.com/shaunluttin/typescript-jagged-array-type-and-prototype
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
Is there a reason we needFlatArray<T>? Can'tflat()'s signature simply beflat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types.[ 1, 'foo' ].flat()should work, no?
– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we returnArray<T>instead ofFlatArray<T>, then resultant type will not be aT. For instance, thenumberexample would return this error at compile time:Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.
– Shaun Luttin
Nov 21 '18 at 19:55
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
3
This is not a complete answer, but it does get us most of the way. Here it is in the TypeScript Playground.
type JaggedArrayItem<T> = T | JaggedArray<T>;
interface JaggedArray<T> extends Array<JaggedArrayItem<T>> { }
type FlatArray<T> = T extends JaggedArrayItem<infer U> ? U : T;
interface Array<T> {
flat(this: JaggedArray<T>): FlatArray<T>;
}
Array.prototype.flat = function () {
return this.reduce(
(arr, val) => Array.isArray(val)
? arr.concat(val.flat())
: arr.concat(val),
);
};
// Test
const myRecursiveArray: JaggedArray<number> = [
10,
[9],
[
[8],
[
[7],
]
],
];
const flattened: number = myRecursiveArray.flat();
const flattenedToo: (string | number) = [1, 'two'].flat();
See also
- https://github.com/Microsoft/TypeScript/issues/3496#issuecomment-128553540
- https://github.com/shaunluttin/typescript-jagged-array-type-and-prototype
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
Is there a reason we needFlatArray<T>? Can'tflat()'s signature simply beflat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types.[ 1, 'foo' ].flat()should work, no?
– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we returnArray<T>instead ofFlatArray<T>, then resultant type will not be aT. For instance, thenumberexample would return this error at compile time:Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.
– Shaun Luttin
Nov 21 '18 at 19:55
add a comment |
3
This is not a complete answer, but it does get us most of the way. Here it is in the TypeScript Playground.
type JaggedArrayItem<T> = T | JaggedArray<T>;
interface JaggedArray<T> extends Array<JaggedArrayItem<T>> { }
type FlatArray<T> = T extends JaggedArrayItem<infer U> ? U : T;
interface Array<T> {
flat(this: JaggedArray<T>): FlatArray<T>;
}
Array.prototype.flat = function () {
return this.reduce(
(arr, val) => Array.isArray(val)
? arr.concat(val.flat())
: arr.concat(val),
);
};
// Test
const myRecursiveArray: JaggedArray<number> = [
10,
[9],
[
[8],
[
[7],
]
],
];
const flattened: number = myRecursiveArray.flat();
const flattenedToo: (string | number) = [1, 'two'].flat();
See also
- https://github.com/Microsoft/TypeScript/issues/3496#issuecomment-128553540
- https://github.com/shaunluttin/typescript-jagged-array-type-and-prototype
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
Is there a reason we needFlatArray<T>? Can'tflat()'s signature simply beflat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types.[ 1, 'foo' ].flat()should work, no?
– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we returnArray<T>instead ofFlatArray<T>, then resultant type will not be aT. For instance, thenumberexample would return this error at compile time:Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.
– Shaun Luttin
Nov 21 '18 at 19:55
add a comment |
3
3
3
This is not a complete answer, but it does get us most of the way. Here it is in the TypeScript Playground.
type JaggedArrayItem<T> = T | JaggedArray<T>;
interface JaggedArray<T> extends Array<JaggedArrayItem<T>> { }
type FlatArray<T> = T extends JaggedArrayItem<infer U> ? U : T;
interface Array<T> {
flat(this: JaggedArray<T>): FlatArray<T>;
}
Array.prototype.flat = function () {
return this.reduce(
(arr, val) => Array.isArray(val)
? arr.concat(val.flat())
: arr.concat(val),
);
};
// Test
const myRecursiveArray: JaggedArray<number> = [
10,
[9],
[
[8],
[
[7],
]
],
];
const flattened: number = myRecursiveArray.flat();
const flattenedToo: (string | number) = [1, 'two'].flat();
See also
- https://github.com/Microsoft/TypeScript/issues/3496#issuecomment-128553540
- https://github.com/shaunluttin/typescript-jagged-array-type-and-prototype
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
This is not a complete answer, but it does get us most of the way. Here it is in the TypeScript Playground.
type JaggedArrayItem<T> = T | JaggedArray<T>;
interface JaggedArray<T> extends Array<JaggedArrayItem<T>> { }
type FlatArray<T> = T extends JaggedArrayItem<infer U> ? U : T;
interface Array<T> {
flat(this: JaggedArray<T>): FlatArray<T>;
}
Array.prototype.flat = function () {
return this.reduce(
(arr, val) => Array.isArray(val)
? arr.concat(val.flat())
: arr.concat(val),
);
};
// Test
const myRecursiveArray: JaggedArray<number> = [
10,
[9],
[
[8],
[
[7],
]
],
];
const flattened: number = myRecursiveArray.flat();
const flattenedToo: (string | number) = [1, 'two'].flat();
See also
- https://github.com/Microsoft/TypeScript/issues/3496#issuecomment-128553540
- https://github.com/shaunluttin/typescript-jagged-array-type-and-prototype
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
edited Nov 21 '18 at 19:51
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
answered Nov 20 '18 at 4:27
Shaun LuttinShaun Luttin
58.1k32223285
58.1k32223285
Is there a reason we needFlatArray<T>? Can'tflat()'s signature simply beflat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types.[ 1, 'foo' ].flat()should work, no?
– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we returnArray<T>instead ofFlatArray<T>, then resultant type will not be aT. For instance, thenumberexample would return this error at compile time:Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.
– Shaun Luttin
Nov 21 '18 at 19:55
add a comment |
Is there a reason we needFlatArray<T>? Can'tflat()'s signature simply beflat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types.[ 1, 'foo' ].flat()should work, no?
– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we returnArray<T>instead ofFlatArray<T>, then resultant type will not be aT. For instance, thenumberexample would return this error at compile time:Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.
– Shaun Luttin
Nov 21 '18 at 19:55
Is there a reason we need
FlatArray<T>? Can't flat()'s signature simply be flat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types. [ 1, 'foo' ].flat() should work, no?– dx_over_dt
Nov 21 '18 at 17:55
Is there a reason we need
FlatArray<T>? Can't flat()'s signature simply be flat(this: JaggedArray<T>): Array<T>;? I'm also unsure of why we need to surface an error for union types. [ 1, 'foo' ].flat() should work, no?– dx_over_dt
Nov 21 '18 at 17:55
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx Good point about the TODO item for surfacing an error. I put that in there because I though that the original question wanted to avoid union types... I think I was mistaken about that, so I deleted that TODO item.
– Shaun Luttin
Nov 21 '18 at 19:50
@dfoverdx If we return
Array<T> instead of FlatArray<T>, then resultant type will not be a T. For instance, the number example would return this error at compile time: Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.– Shaun Luttin
Nov 21 '18 at 19:55
@dfoverdx If we return
Array<T> instead of FlatArray<T>, then resultant type will not be a T. For instance, the number example would return this error at compile time: Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArrayItem<number>' is not assignable to type 'number'. Type 'JaggedArray<number>' is not assignable to type 'number'.– Shaun Luttin
Nov 21 '18 at 19:55
add a comment |
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