Reputation: 4402
How do I type a function, so that the input object is the same as the output object, but with different values?
//a has type { a: number;b: number }
let a = { 'a': 1, 'b': 1 };
interface IDictNumber {
[key: string]: number;
}
interface IDictString {
[key: string]: string;
}
function convert(f: IDictNumber) {
return Object.keys(f)
.reduce((p, v) => {
p[v] = `${f[v]}`;
return p;
},
{} as IDictString);
}
//b has type IDictString, but I wanted it to have { a: string;b: string }
let b= convert(a);
Upvotes: 0
Views: 158
Reputation: 4402
This is now possible in TypeScript 2.1:
//a has type { a: number;b: number }
let a = { 'a': 1, 'b': 1 };
type Convert<T,K> = {
[P in keyof T]: K;
};
function convertToString<T>(f: T) {
let a = <(keyof T)[]>Object.keys(f);
return a.reduce((p, v) => {
p[v] = `${f[v]}`;
return p;
},
{} as Convert<T, string>);
}
let b = convertToString(a);
Upvotes: 2
Reputation: 37938
You could use generics for this:
let a = { 'a': 1, 'b': 1 };
interface A<T extends string|number> {
[key: string]: T;
a: T,
b: T
}
function convert(f: A<number>) {
return Object.keys(f)
.reduce((p, v) => {
p[v] = `${f[v]}`;
return p;
},
{} as A<string>);
}
let b = convert(a);
Upvotes: 0