Leveraging both the spread operator and optional fields can improve the productivity and readability of your

Question

Leveraging both the spread operator and optional fields can improve the productivity and readability of your

Imagine you have an object with a mandatory field that cannot be null:

interface MyTypeMandatory { 
    value: number;
}

Now, you want to update this object using fields from another object, but this time with an optional field:

interface MyTypeOptional { 
    value?: number;
}

So, you decide to create a function for this task:

function mergeObjects(a: MyTypeMandatory, b: MyTypeOptional) {
    return { ...a, ...b };
}

The question arises - what will be the expected return type of this function?

const result = mergeObjects({ value: 1 }, { value: undefined });

Experimenting reveals that it seems to adhere to the MyTypeMandatory interface, even when one of the spreads includes an optional field.

Interestingly, switching the order of the spreads doesn't affect the inferred type, despite potentially changing the actual runtime type.

function mergeObjects(a: MyTypeMandatory, b: MyTypeOptional) {
    return { ...b, ...a };
}

Why does TypeScript exhibit such behavior and what are some ways to navigate around this particular issue?

typescript spread-syntax

Answer 1

Answer №1

When dealing with object spread types, certain rules are applied as outlined in this guide:

Call and construct signatures are removed, only non-method properties are retained, and in cases where the same property name exists, the type of the rightmost property takes precedence.

With object literals containing generic spread expressions, intersection types are now generated, akin to the behavior of the Object.assign function and JSX literals. For instance:

Everything seems to work smoothly until an optional property is present in the rightmost argument. The ambiguity arises when dealing with a property like {value?: number; }, which could signify either a missing property or a property set to undefined. TypeScript struggles to differentiate between these two scenarios using the optional modifier notation ?. Let's consider an example:

const t1: { a: number } = { a: 3 }
const u1: { a?: string } = { a: undefined }

const spread1 = { ...u1 } // { a?: string | undefined; }
const spread2 = { ...t1, ...u1 } // { a: string | number; }
const spread3 = { ...u1, ...t1 } // { a: number; }

spread1

makes sense - the key a can be defined or left undefined. In this case, we must use the undefined type to represent the absence of a property value.

spread2

The type of a is dictated by the rightmost argument. If a is present in u1, it would be of type string; otherwise, the spread operation retrieves the a property from the first argument t1, which has a type of number. Thus, string | number is a reasonable outcome in this context. Note that there is no mention of undefined here because TypeScript assumes that the property does not exist at all or that it is a string. To observe a different result, we could assign an explicit property value type of undefined to a:

const u2 = { a: undefined }
const spread4 = { ...t1, ...u2 } // { a: undefined; }

spread3

In this scenario, the value of a from t1 replaces the value of a from u1, resulting in a return type of number.

I wouldn't anticipate an immediate resolution to this issue based on discussions. Therefore, a potential workaround involves introducing a distinct Spread type and function:

type Spread<L, R> = Pick<L, Exclude<keyof L, keyof R>> & R;

function spread<T, U>(a: T, b: U): Spread<T, U> {
    return { ...a, ...b };
}

const t1_: { a: number; b: string }
const t2_: { a?: number; b: string }
const u1_: { a: number; c: boolean }
const u2_: { a?: number; c: boolean }

const t1u2 = spread(t1_, u2_); // { b: string; a?: number | undefined; c: boolean; }
const t2u1 = spread(t2_, u1_); // { b: string; a: number; c: boolean; }

Hoping this provides some clarity! Check out this interactive demo for the above code.

Answer 2

When dealing with object spread types, certain rules are applied as outlined in this guide:

Call and construct signatures are removed, only non-method properties are retained, and in cases where the same property name exists, the type of the rightmost property takes precedence.

With object literals containing generic spread expressions, intersection types are now generated, akin to the behavior of the Object.assign function and JSX literals. For instance:

Everything seems to work smoothly until an optional property is present in the rightmost argument. The ambiguity arises when dealing with a property like {value?: number; }, which could signify either a missing property or a property set to undefined. TypeScript struggles to differentiate between these two scenarios using the optional modifier notation ?. Let's consider an example:

const t1: { a: number } = { a: 3 }
const u1: { a?: string } = { a: undefined }

const spread1 = { ...u1 } // { a?: string | undefined; }
const spread2 = { ...t1, ...u1 } // { a: string | number; }
const spread3 = { ...u1, ...t1 } // { a: number; }

spread1

makes sense - the key a can be defined or left undefined. In this case, we must use the undefined type to represent the absence of a property value.

spread2

The type of a is dictated by the rightmost argument. If a is present in u1, it would be of type string; otherwise, the spread operation retrieves the a property from the first argument t1, which has a type of number. Thus, string | number is a reasonable outcome in this context. Note that there is no mention of undefined here because TypeScript assumes that the property does not exist at all or that it is a string. To observe a different result, we could assign an explicit property value type of undefined to a:

const u2 = { a: undefined }
const spread4 = { ...t1, ...u2 } // { a: undefined; }

spread3

In this scenario, the value of a from t1 replaces the value of a from u1, resulting in a return type of number.

I wouldn't anticipate an immediate resolution to this issue based on discussions. Therefore, a potential workaround involves introducing a distinct Spread type and function:

type Spread<L, R> = Pick<L, Exclude<keyof L, keyof R>> & R;

function spread<T, U>(a: T, b: U): Spread<T, U> {
    return { ...a, ...b };
}

const t1_: { a: number; b: string }
const t2_: { a?: number; b: string }
const u1_: { a: number; c: boolean }
const u2_: { a?: number; c: boolean }

const t1u2 = spread(t1_, u2_); // { b: string; a?: number | undefined; c: boolean; }
const t2u1 = spread(t2_, u1_); // { b: string; a: number; c: boolean; }

Hoping this provides some clarity! Check out this interactive demo for the above code.

Answer 3

Answer №2

Disclaimer: The following is my personal interpretation and theory about the subject, as it has not been confirmed.

An interesting issue arises when dealing with optional fields in TypeScript.

Consider defining a type like this:

interface MyTypeOptional { 
    value?: number;
}

In this case, TypeScript expects the value to be of type number | never when spreading the object.

However, due to the ambiguous nature of optional fields versus undefined values, TypeScript sometimes infers the spread object with an optional field type of number | undefined instead of number | never.

Strangely, when spreading a non-optional type alongside an optional type, TypeScript correctly casts the optional field into number | never.

To work around this inconsistency, it's advisable to avoid using optional fields until the TypeScript team resolves this issue. This can impact how you handle object keys being omitted if undefined, the usage of Partial types, and handling unset values within your application.

For more information on this limitation of optional fields in TypeScript, refer to the open issue here: https://github.com/microsoft/TypeScript/issues/13195

Answer 4

Disclaimer: The following is my personal interpretation and theory about the subject, as it has not been confirmed.

An interesting issue arises when dealing with optional fields in TypeScript.

Consider defining a type like this:

interface MyTypeOptional { 
    value?: number;
}

In this case, TypeScript expects the value to be of type number | never when spreading the object.

However, due to the ambiguous nature of optional fields versus undefined values, TypeScript sometimes infers the spread object with an optional field type of number | undefined instead of number | never.

Strangely, when spreading a non-optional type alongside an optional type, TypeScript correctly casts the optional field into number | never.

To work around this inconsistency, it's advisable to avoid using optional fields until the TypeScript team resolves this issue. This can impact how you handle object keys being omitted if undefined, the usage of Partial types, and handling unset values within your application.

For more information on this limitation of optional fields in TypeScript, refer to the open issue here: https://github.com/microsoft/TypeScript/issues/13195

Answer 5

Answer №3

To address this issue, one possible solution is to utilize type level function as demonstrated below:

interface MyTypeRequired { 
    a: number;
    value: number;
}

interface MyTypeOptional { 
    b: string;
    value?: number;
}

type MergedWithOptional<T1, T2> = {
    [K in keyof T1]: K extends keyof T2 ? T2[K] extends undefined ? undefined : T2[K] : T1[K] 
} & {
    [K in Exclude<keyof T2, keyof T1>]: T2[K]
}

function createObject(a: MyTypeRequired, b: MyTypeOptional): MergedWithOptional<MyTypeRequired, MyTypeOptional> {
    return { ...b, ...a };
}

For testing purposes, additional fields have been included to observe the behavior of the solution. The essence lies in augmenting the result with T1[K] | undefined when encountering optional fields with possible undefined values in the second object. Meanwhile, the merger includes all other fields unique to T2 compared to T1.

Further explanations include:

```
K extends keyof T2 ? T2[K] extends undefined ? undefined : T2[K] : T1[K]
```
conditions the addition of undefined to the value type if the key exists in the second object and can be undefined, demonstrating the behavior of conditional types for union types.
```
Exclude<keyof T2, keyof T1>
```
- selects only the keys that are exclusive to the second object.

Answer 6

To address this issue, one possible solution is to utilize type level function as demonstrated below:

interface MyTypeRequired { 
    a: number;
    value: number;
}

interface MyTypeOptional { 
    b: string;
    value?: number;
}

type MergedWithOptional<T1, T2> = {
    [K in keyof T1]: K extends keyof T2 ? T2[K] extends undefined ? undefined : T2[K] : T1[K] 
} & {
    [K in Exclude<keyof T2, keyof T1>]: T2[K]
}

function createObject(a: MyTypeRequired, b: MyTypeOptional): MergedWithOptional<MyTypeRequired, MyTypeOptional> {
    return { ...b, ...a };
}

For testing purposes, additional fields have been included to observe the behavior of the solution. The essence lies in augmenting the result with T1[K] | undefined when encountering optional fields with possible undefined values in the second object. Meanwhile, the merger includes all other fields unique to T2 compared to T1.

Further explanations include:

```
K extends keyof T2 ? T2[K] extends undefined ? undefined : T2[K] : T1[K]
```
conditions the addition of undefined to the value type if the key exists in the second object and can be undefined, demonstrating the behavior of conditional types for union types.
```
Exclude<keyof T2, keyof T1>
```
- selects only the keys that are exclusive to the second object.

Leveraging both the spread operator and optional fields can improve the productivity and readability of your

Answer №1

Answer №2

Answer №3

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