Preserving type information in TypeScript function return values

Question

Preserving type information in TypeScript function return values

Wondering how to make this code work in TypeScript. Function tempA copies the value of x.a to x.temp and returns it, while function tempB does the same with x.b. However, when calling tempA, the compiler seems to forget about the existence of the b field on x. Is there a workaround for this issue? Is there a GitHub problem addressing this scenario or which feature category should it be reported under?

let tempA = (x: { a: number; temp?: number}) => {
    x.temp = x.a;
    return x;
}

let tempB = (x: { b: number; temp?: number}) => {
    x.temp = x.b
    return x;
}

let x = { a: 4, b: 3 };
let y = tempA(x);
let z = tempB(y); // Error! y.b is considered non-existent by the compiler (even though it exists)

typescript

Answer 1

Answer №1

Given that the parameter type of x is defined as { a: number; temp?: number }, and the function returns x, the inferred return type becomes { a: number, temp?: number }. Consequently, this is also the type of y, causing Typescript to be unaware of the existence of the property b.

To address this issue, you can opt for making the function generic. This approach ensures that it will always return something with the same structure as its input:

function tempA<T extends { a: number; temp?: number }>(x: T): T {
    x.temp = x.a;
    return x;
}

This implementation is satisfactory for your current scenario but isn't as rigorous as it could be. The return type implies that the object may not include a temp property, despite the assurance that it does. To enforce a stricter return type, an intersection & can be utilized, albeit adding complexity which might not align with your specific needs:

function tempA<T extends { a: number; temp?: number }>(x: T): T & { temp: number }
{
    x.temp = x.a;
    return x as T & { temp: number };
}

Playground Link

Answer 2

Given that the parameter type of x is defined as { a: number; temp?: number }, and the function returns x, the inferred return type becomes { a: number, temp?: number }. Consequently, this is also the type of y, causing Typescript to be unaware of the existence of the property b.

To address this issue, you can opt for making the function generic. This approach ensures that it will always return something with the same structure as its input:

function tempA<T extends { a: number; temp?: number }>(x: T): T {
    x.temp = x.a;
    return x;
}

This implementation is satisfactory for your current scenario but isn't as rigorous as it could be. The return type implies that the object may not include a temp property, despite the assurance that it does. To enforce a stricter return type, an intersection & can be utilized, albeit adding complexity which might not align with your specific needs:

function tempA<T extends { a: number; temp?: number }>(x: T): T & { temp: number }
{
    x.temp = x.a;
    return x as T & { temp: number };
}

Playground Link

Answer 3

Answer №2

Here, the variable x is defined with a type signature that consists of properties 'a' and 'b', both being numbers.

let x: {a: number, b: number} = {a: 4, b: 3}

Next, you pass this variable into a method called tempA, which has its own type signature as follows:

(x: { a: number; temp?: number }) => { a: number; temp?: number }

The issue arises when you try to assign the result of tempA to another variable 'y' like this:

let y: { a: number; temp?: number } = tempA(x);

It's apparent that the property 'b' is missing in this case. This behavior is not a bug but rather an intentional design choice. A similar scenario is illustrated here:

let y = tempA({ a: 4 });
// Ignore compiler warnings
// @ts-ignore
let z: { b: number; temp?: number} = tempB(y) 
// Assignment to z.b results in undefined, violating type constraints

To address this issue, you'll need to rethink and update your type signatures accordingly.

// Modify the type signature of tempA to ensure any 'x' passed must have a 'b' property
let tempA = (x: { a: number; b: number; temp?: number}) => {
    x.temp = x.a;
    return x;
}

let tempB = (x: { b: number; temp?: number}) => {
    x.temp = x.b
    return x;
}

let x = { a: 4, b: 3 };
let y = tempA(x);
let z = tempB(y); // Resolves the issue

Answer 4

Here, the variable x is defined with a type signature that consists of properties 'a' and 'b', both being numbers.

let x: {a: number, b: number} = {a: 4, b: 3}

Next, you pass this variable into a method called tempA, which has its own type signature as follows:

(x: { a: number; temp?: number }) => { a: number; temp?: number }

The issue arises when you try to assign the result of tempA to another variable 'y' like this:

let y: { a: number; temp?: number } = tempA(x);

It's apparent that the property 'b' is missing in this case. This behavior is not a bug but rather an intentional design choice. A similar scenario is illustrated here:

let y = tempA({ a: 4 });
// Ignore compiler warnings
// @ts-ignore
let z: { b: number; temp?: number} = tempB(y) 
// Assignment to z.b results in undefined, violating type constraints

To address this issue, you'll need to rethink and update your type signatures accordingly.

// Modify the type signature of tempA to ensure any 'x' passed must have a 'b' property
let tempA = (x: { a: number; b: number; temp?: number}) => {
    x.temp = x.a;
    return x;
}

let tempB = (x: { b: number; temp?: number}) => {
    x.temp = x.b
    return x;
}

let x = { a: 4, b: 3 };
let y = tempA(x);
let z = tempB(y); // Resolves the issue

Preserving type information in TypeScript function return values

Answer №1

Answer №2

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