Typescript base class reference subclass
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0
down vote
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I am trying to create a tree-like structure with inheritance like the following:
base.ts:
export class BaseClass {
children: BaseClass = ;
get subclassChildren(): SubClass {
return this.children.filter((child): child is SubClass => child instanceof SubClass);
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {}
This wont work because the base class cannot find SubClass:
TSError: ⨯ Unable to compile TypeScript:
base.ts(4,26): error TS2304: Cannot find name 'SubClass'.
base.ts(5,49): error TS2304: Cannot find name 'SubClass'.
base.ts(5,78): error TS2304: Cannot find name 'SubClass'.
Not unexpected but if I try to import the subclass in base.ts it wont work either as there are circular references to each other:
TypeError: Object prototype may only be an Object or null: undefined
Is there a way to forward declare SubClass or some other way to get this code working?
EDIT:
tsconfig.json (generated by tsc --init):
{
"compilerOptions": {
"target": "es5",
"module": "commonjs",
"strict": true,
"esModuleInterop": true
}
}
typescript
add a comment |
up vote
0
down vote
favorite
I am trying to create a tree-like structure with inheritance like the following:
base.ts:
export class BaseClass {
children: BaseClass = ;
get subclassChildren(): SubClass {
return this.children.filter((child): child is SubClass => child instanceof SubClass);
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {}
This wont work because the base class cannot find SubClass:
TSError: ⨯ Unable to compile TypeScript:
base.ts(4,26): error TS2304: Cannot find name 'SubClass'.
base.ts(5,49): error TS2304: Cannot find name 'SubClass'.
base.ts(5,78): error TS2304: Cannot find name 'SubClass'.
Not unexpected but if I try to import the subclass in base.ts it wont work either as there are circular references to each other:
TypeError: Object prototype may only be an Object or null: undefined
Is there a way to forward declare SubClass or some other way to get this code working?
EDIT:
tsconfig.json (generated by tsc --init):
{
"compilerOptions": {
"target": "es5",
"module": "commonjs",
"strict": true,
"esModuleInterop": true
}
}
typescript
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to create a tree-like structure with inheritance like the following:
base.ts:
export class BaseClass {
children: BaseClass = ;
get subclassChildren(): SubClass {
return this.children.filter((child): child is SubClass => child instanceof SubClass);
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {}
This wont work because the base class cannot find SubClass:
TSError: ⨯ Unable to compile TypeScript:
base.ts(4,26): error TS2304: Cannot find name 'SubClass'.
base.ts(5,49): error TS2304: Cannot find name 'SubClass'.
base.ts(5,78): error TS2304: Cannot find name 'SubClass'.
Not unexpected but if I try to import the subclass in base.ts it wont work either as there are circular references to each other:
TypeError: Object prototype may only be an Object or null: undefined
Is there a way to forward declare SubClass or some other way to get this code working?
EDIT:
tsconfig.json (generated by tsc --init):
{
"compilerOptions": {
"target": "es5",
"module": "commonjs",
"strict": true,
"esModuleInterop": true
}
}
typescript
I am trying to create a tree-like structure with inheritance like the following:
base.ts:
export class BaseClass {
children: BaseClass = ;
get subclassChildren(): SubClass {
return this.children.filter((child): child is SubClass => child instanceof SubClass);
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {}
This wont work because the base class cannot find SubClass:
TSError: ⨯ Unable to compile TypeScript:
base.ts(4,26): error TS2304: Cannot find name 'SubClass'.
base.ts(5,49): error TS2304: Cannot find name 'SubClass'.
base.ts(5,78): error TS2304: Cannot find name 'SubClass'.
Not unexpected but if I try to import the subclass in base.ts it wont work either as there are circular references to each other:
TypeError: Object prototype may only be an Object or null: undefined
Is there a way to forward declare SubClass or some other way to get this code working?
EDIT:
tsconfig.json (generated by tsc --init):
{
"compilerOptions": {
"target": "es5",
"module": "commonjs",
"strict": true,
"esModuleInterop": true
}
}
typescript
typescript
edited Nov 20 at 13:53
asked Nov 20 at 13:22
ext
75821227
75821227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
Try this:
base.ts:
export class BaseClass {
className = "base"
children: BaseClass = ;
get subclassChildren() {
return this.children.filter((child) => { return child.className == "subClass" });
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
className = "subClass"
}
The major issue here is thatsubclassChildrenis not available onBaseClassany longer. And sincechildrenis still an array ofBaseClassit would need a lot of casting.
– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer aSubClassbut aBaseClass.
– ext
Nov 20 at 13:58
add a comment |
up vote
0
down vote
accepted
Using module augmentation it is possible to add methods and properties to existing classes:
base.ts:
export class BaseClass {
children: BaseClass = ;
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
foo(){
console.log('foo');
}
}
/* augment base class */
declare module "./base" {
interface BaseClass {
subclassChildren: SubClass;
bar(): void;
}
}
/* implement getter */
Object.defineProperty(BaseClass.prototype, 'subclassChildren', {
get(){
return this.children.filter((child: BaseClass): child is SubClass => child instanceof SubClass);
},
});
/* implement method */
BaseClass.prototype.bar = function bar(){
console.log('bar');
}
When using it both base.ts and sub.ts has to be imported as only sub.ts contains the augmented class:
import { BaseClass } from './base';
import { SubClass } from './sub';
const root = new BaseClass();
root.children.push(new BaseClass());
root.children.push(new SubClass());
root.subclassChildren.map((child) => child.foo());
root.bar();
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try this:
base.ts:
export class BaseClass {
className = "base"
children: BaseClass = ;
get subclassChildren() {
return this.children.filter((child) => { return child.className == "subClass" });
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
className = "subClass"
}
The major issue here is thatsubclassChildrenis not available onBaseClassany longer. And sincechildrenis still an array ofBaseClassit would need a lot of casting.
– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer aSubClassbut aBaseClass.
– ext
Nov 20 at 13:58
add a comment |
up vote
0
down vote
Try this:
base.ts:
export class BaseClass {
className = "base"
children: BaseClass = ;
get subclassChildren() {
return this.children.filter((child) => { return child.className == "subClass" });
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
className = "subClass"
}
The major issue here is thatsubclassChildrenis not available onBaseClassany longer. And sincechildrenis still an array ofBaseClassit would need a lot of casting.
– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer aSubClassbut aBaseClass.
– ext
Nov 20 at 13:58
add a comment |
up vote
0
down vote
up vote
0
down vote
Try this:
base.ts:
export class BaseClass {
className = "base"
children: BaseClass = ;
get subclassChildren() {
return this.children.filter((child) => { return child.className == "subClass" });
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
className = "subClass"
}
Try this:
base.ts:
export class BaseClass {
className = "base"
children: BaseClass = ;
get subclassChildren() {
return this.children.filter((child) => { return child.className == "subClass" });
}
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
className = "subClass"
}
edited Nov 20 at 13:54
answered Nov 20 at 13:35
javimovi
318110
318110
The major issue here is thatsubclassChildrenis not available onBaseClassany longer. And sincechildrenis still an array ofBaseClassit would need a lot of casting.
– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer aSubClassbut aBaseClass.
– ext
Nov 20 at 13:58
add a comment |
The major issue here is thatsubclassChildrenis not available onBaseClassany longer. And sincechildrenis still an array ofBaseClassit would need a lot of casting.
– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer aSubClassbut aBaseClass.
– ext
Nov 20 at 13:58
The major issue here is that
subclassChildren is not available on BaseClass any longer. And since children is still an array of BaseClass it would need a lot of casting.– ext
Nov 20 at 13:39
The major issue here is that
subclassChildren is not available on BaseClass any longer. And since children is still an array of BaseClass it would need a lot of casting.– ext
Nov 20 at 13:39
And in this way?
– javimovi
Nov 20 at 13:49
And in this way?
– javimovi
Nov 20 at 13:49
This will filter out the subclasses but not cast them any longer, what I get is no longer a
SubClass but a BaseClass.– ext
Nov 20 at 13:58
This will filter out the subclasses but not cast them any longer, what I get is no longer a
SubClass but a BaseClass.– ext
Nov 20 at 13:58
add a comment |
up vote
0
down vote
accepted
Using module augmentation it is possible to add methods and properties to existing classes:
base.ts:
export class BaseClass {
children: BaseClass = ;
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
foo(){
console.log('foo');
}
}
/* augment base class */
declare module "./base" {
interface BaseClass {
subclassChildren: SubClass;
bar(): void;
}
}
/* implement getter */
Object.defineProperty(BaseClass.prototype, 'subclassChildren', {
get(){
return this.children.filter((child: BaseClass): child is SubClass => child instanceof SubClass);
},
});
/* implement method */
BaseClass.prototype.bar = function bar(){
console.log('bar');
}
When using it both base.ts and sub.ts has to be imported as only sub.ts contains the augmented class:
import { BaseClass } from './base';
import { SubClass } from './sub';
const root = new BaseClass();
root.children.push(new BaseClass());
root.children.push(new SubClass());
root.subclassChildren.map((child) => child.foo());
root.bar();
add a comment |
up vote
0
down vote
accepted
Using module augmentation it is possible to add methods and properties to existing classes:
base.ts:
export class BaseClass {
children: BaseClass = ;
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
foo(){
console.log('foo');
}
}
/* augment base class */
declare module "./base" {
interface BaseClass {
subclassChildren: SubClass;
bar(): void;
}
}
/* implement getter */
Object.defineProperty(BaseClass.prototype, 'subclassChildren', {
get(){
return this.children.filter((child: BaseClass): child is SubClass => child instanceof SubClass);
},
});
/* implement method */
BaseClass.prototype.bar = function bar(){
console.log('bar');
}
When using it both base.ts and sub.ts has to be imported as only sub.ts contains the augmented class:
import { BaseClass } from './base';
import { SubClass } from './sub';
const root = new BaseClass();
root.children.push(new BaseClass());
root.children.push(new SubClass());
root.subclassChildren.map((child) => child.foo());
root.bar();
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Using module augmentation it is possible to add methods and properties to existing classes:
base.ts:
export class BaseClass {
children: BaseClass = ;
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
foo(){
console.log('foo');
}
}
/* augment base class */
declare module "./base" {
interface BaseClass {
subclassChildren: SubClass;
bar(): void;
}
}
/* implement getter */
Object.defineProperty(BaseClass.prototype, 'subclassChildren', {
get(){
return this.children.filter((child: BaseClass): child is SubClass => child instanceof SubClass);
},
});
/* implement method */
BaseClass.prototype.bar = function bar(){
console.log('bar');
}
When using it both base.ts and sub.ts has to be imported as only sub.ts contains the augmented class:
import { BaseClass } from './base';
import { SubClass } from './sub';
const root = new BaseClass();
root.children.push(new BaseClass());
root.children.push(new SubClass());
root.subclassChildren.map((child) => child.foo());
root.bar();
Using module augmentation it is possible to add methods and properties to existing classes:
base.ts:
export class BaseClass {
children: BaseClass = ;
}
sub.ts:
import { BaseClass } from './base';
export class SubClass extends BaseClass {
foo(){
console.log('foo');
}
}
/* augment base class */
declare module "./base" {
interface BaseClass {
subclassChildren: SubClass;
bar(): void;
}
}
/* implement getter */
Object.defineProperty(BaseClass.prototype, 'subclassChildren', {
get(){
return this.children.filter((child: BaseClass): child is SubClass => child instanceof SubClass);
},
});
/* implement method */
BaseClass.prototype.bar = function bar(){
console.log('bar');
}
When using it both base.ts and sub.ts has to be imported as only sub.ts contains the augmented class:
import { BaseClass } from './base';
import { SubClass } from './sub';
const root = new BaseClass();
root.children.push(new BaseClass());
root.children.push(new SubClass());
root.subclassChildren.map((child) => child.foo());
root.bar();
answered Nov 23 at 16:33
ext
75821227
75821227
add a comment |
add a comment |
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