typescript generic function return type from paramaters-CodePudding

I want dynamic return type from parameters.

type TestType = {
    value1: string,
    value2: number
}

function testFn1<T extend TestType>(...pick:(keyof T)[]): ??? { }

//testFn<TestType>("value1")  expect {value1: string}
//testFn<TestType>("value1","value2") expect {value1: string, value2:number}

Plz

How can i define function return type?

I try below.

function test3<T, K extends keyof T>(...vars: K[]): Record<K, string | number>

but this return Just T

CodePudding user response：

For the return type, you can use the Pick utility type.

But what you also need is to actually pass a T to the function. One, so the function actually has an object to pick values from, and two, so the compiler can infer T.

With that, this works:

function test<T, K extends keyof T>(t: T, ...vars: K[]): Pick<T, K> {
  // ...
}

const t: TestType = fromSomewhere();

const result = test(t, "value1", "value2");
console.log(result.value1);
console.log(result.value2)

const result2 = test(t, "value1");
console.log(result2.value1);
console.log(result2.value2); // error, as expected

CodePudding user response：

TypeScript does not have partial type inference. If you provide TestType explicitly to T, it can not infer other generic types via argument types.

One work-around would be to use currying. You can explicitly provide TestType as the generic type in the first function call while the second call infers K with the arguments.

function testFn<T extends TestType>() {
    return <K extends keyof T>(...pick: K[]): { [Key in K]: T[Key] } => {
        return null!
    }
}

const result1 = testFn<TestType>()("value1")
const result2 = testFn<TestType>()("value1","value2")

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