for example, I have an interface:
interface Foo {
value1: number;
value2: number;
}
I'd like to add a sum() method that would give me a sum of values on any instance of this interface. What is a Typescript's way of doing this?
I need to be able to write a function like this:
function printFooSum(foo: Foo) {
console.log(???);
}
and the function should not involve manual assignment of foo properties to some other class.
Extend a TypeScript interface via the extends keyword:
interface Foo {
value1: number;
value2: number;
}
interface Bar extends Foo {
sum() : number;
}
Implement the interface the following way:
class Baz implements Bar {
public value1: number;
public value2: number;
sum() : number { return this.value1 + this.value2; }
}
var bar : Bar = new Baz();
Note: the implements Bar part is optional. Instances of Baz would still be assignable to Bar slots, as they have the required properties. However adding implements Bar will enforce the compiler to check the Baz class implements the Bar interface, throwing a compile time error on Baz if it doesn't.
One way to do it is the following:
interface Foo {
value1: number;
value2: number;
}
class Foo {
public static sum(foo: Foo) {
return foo.value1 + foo.value2;
}
}
function printFooSum(foo: Foo) {
console.log(Foo.sum(foo));
}
But maybe there is a more elegant way...
I think you can do something like:
interface Foo1 extends Foo {
suma(): number
}
interface Foo{
value1: number;
value2: number;
}
http://www.typescriptlang.org/docs/handbook/interfaces.html#extending-interfaces
You add this:
I need to be able to write a function like this:
function printFooSum(foo: Foo) {
console.log(???);
}
But this part does not understand well:
and the function should not involve manual assignment of foo properties to some other class.
you do not want to do this?
function printFooSum(foo: Foo) {
console.log(foo.value1 + foo.value2);
}
