4 replies

Typescript "implements"?

Hello, guys, i have two interfaces that have same name fields but different implementation, but i want when i create new interface without all fields from A in new interface, it should throws an error any help please?

interface A{
a: boolean;
b: boolean;
}
interface B{
a: function;
b:function
}

interface A{
a: boolean;
b: boolean;
}
interface B{
a: function;
b:function
}

Solution

What you're asking for is almost like an

interface B implements A {}

interface B implements A {}

- which doesn't work.

I tried a few different things. I'm not sure there's a way to do exactly what you're asking.

The closest I can get is a single interface with unknown types, where the class can then define the types however it wants

interface Stub {
  foo: unknown
  bar: unknown
}

class MyStub implements Stub {
  foo = 'a'
  bar() { return this.foo }
}

interface Stub {
  foo: unknown
  bar: unknown
}

class MyStub implements Stub {
  foo = 'a'
  bar() { return this.foo }
}

And if

baz: unknown

baz: unknown

is added to the interface then

MyStub

MyStub

will have an error that it needs it added as well

You could then extend those classes another layer to get close to what you're asking for, but keys that are not re-defined will inherit from their parents. I'm assuming this is also not ideal.

class MyStub implements Stub {
  foo = ''
  bar = ''
}

class MySpecialStub extends MyStub {
  foo = 'hi'
  // bar = 'bye' // optional, if not defined it's empty string from MyStub
}

class MyStub implements Stub {
  foo = ''
  bar = ''
}

class MySpecialStub extends MyStub {
  foo = 'hi'
  // bar = 'bye' // optional, if not defined it's empty string from MyStub
}

---

You can also do something similar with generics

type A = {
  a: unknown
  b: unknown
}

type B<T extends A> = {}

type C = B<{ a: string }> // error: Property 'b' is missing

type A = {
  a: unknown
  b: unknown
}

type B<T extends A> = {}

type C = B<{ a: string }> // error: Property 'b' is missing

but then you lose

implements

implements

interface D extends B<{ a: string, b: string }> {
  c: unknown
}

class MyThing implements D {} // errors that 'c' is missing, but does not error about 'a' or 'b'

interface D extends B<{ a: string, b: string }> {
  c: unknown
}

class MyThing implements D {} // errors that 'c' is missing, but does not error about 'a' or 'b'

---

Maybe someone else will get closer, but if you explain more of your exact scenario there might also be a better pattern for it.

Using runtime errors rather than TS ones could certainly handle this if that's a better compromise. Other languages have virtual classes that get close to this too. I personally avoid classes. Using the types and then defining objects that conform to those types would get you there if you can give up on

implements

implements

and having classes.

Jump to solution

TTC

Theo's Typesafe Cult•2y ago•

4 replies

choco

Typescript "implements"?

interface A{
a: boolean;
b: boolean;
}
interface B{
a: function;
b:function
}

interface A{
a: boolean;
b: boolean;
}
interface B{
a: function;
b:function
}

Solution

What you're asking for is almost like an

interface B implements A {}

interface B implements A {}

interface Stub {
  foo: unknown
  bar: unknown
}

class MyStub implements Stub {
  foo = 'a'
  bar() { return this.foo }
}

interface Stub {
  foo: unknown
  bar: unknown
}

class MyStub implements Stub {
  foo = 'a'
  bar() { return this.foo }
}

And if

baz: unknown

baz: unknown

is added to the interface then

MyStub

MyStub

class MyStub implements Stub {
  foo = ''
  bar = ''
}

class MySpecialStub extends MyStub {
  foo = 'hi'
  // bar = 'bye' // optional, if not defined it's empty string from MyStub
}

class MyStub implements Stub {
  foo = ''
  bar = ''
}

class MySpecialStub extends MyStub {
  foo = 'hi'
  // bar = 'bye' // optional, if not defined it's empty string from MyStub
}

---

You can also do something similar with generics

type A = {
  a: unknown
  b: unknown
}

type B<T extends A> = {}

type C = B<{ a: string }> // error: Property 'b' is missing

type A = {
  a: unknown
  b: unknown
}

type B<T extends A> = {}

type C = B<{ a: string }> // error: Property 'b' is missing

but then you lose

implements

implements

interface D extends B<{ a: string, b: string }> {
  c: unknown
}

class MyThing implements D {} // errors that 'c' is missing, but does not error about 'a' or 'b'

interface D extends B<{ a: string, b: string }> {
  c: unknown
}

class MyThing implements D {} // errors that 'c' is missing, but does not error about 'a' or 'b'

implements

implements

and having classes.

Jump to solution

Typescript "implements"?

Similar Threads

Typescript "implements"?

Similar Threads

Similar Threads

Similar Threads