What is a way to construct an object without resorting to casts or manually declaring variables for each field?

Question

What is a way to construct an object without resorting to casts or manually declaring variables for each field?

Click here for a hands-on example. Take a look at the code snippet below:

export type BigType = {
  foo: number;  
  bar?: number;
  baz: number;
  qux?: string[];
};

function BuildBigType(params: string[]) {
    // Here's what I'd like to do:
    const endResult: BigType = {};

    //some logic for getting foo
    endResult.foo = 5;

    //some logic for getting bar
    endResult.bar = undefined;

    //...

    return endResult;
}

The issue lies in the fact that {} doesn't conform to the structure of BigType. Two potential solutions come to mind:

Type cast endResult to any and proceed with assigning the fields as demonstrated in the example.
Declare local variables for each field in BigType, then assign them accordingly at the end of the function. For example:
```
  //...
  return {foo, bar, baz, qux};
}
```

Are there any alternative approaches worth considering?

typescript

Answer 1

Answer №1

Finding a completely elegant solution may be challenging. My suggestion would be to go with option 2., as type inferences tend to work well when used together, and you can wait until the end to assemble the full type when you return. To further solidify this, you can explicitly set the return type of your function to avoid any potential errors.

function BuildBigType(params: string[]): BigType {
    //logic for obtaining foo
    const foo = 5;

    //logic for obtaining bar
    const bar = undefined;

    //...

    return {foo, bar, baz, qux};
}

Answer 2

Finding a completely elegant solution may be challenging. My suggestion would be to go with option 2., as type inferences tend to work well when used together, and you can wait until the end to assemble the full type when you return. To further solidify this, you can explicitly set the return type of your function to avoid any potential errors.

function BuildBigType(params: string[]): BigType {
    //logic for obtaining foo
    const foo = 5;

    //logic for obtaining bar
    const bar = undefined;

    //...

    return {foo, bar, baz, qux};
}

Answer 3

Answer №2

If you're searching for a way to implement lazy object initialization in TypeScript, you may be interested to know that it is not directly supported. You can check out the open feature request microsoft/TypeScript#28503 and show your support by giving it a thumbs up.

One way to modify your code to avoid errors without resorting to using the potentially unsafe `any` type is to utilize a type assertion when declaring the variable. This would look like:

function BuildBigTypeAssert(params: string[]) {
  const endResult = {} as BigType;
  endResult.foo = 5;
  endResult.bar = undefined;
  return endResult; // no errors, but be cautious
}

While this approach is acceptable, keep in mind that the compiler won't catch an error if you forget to initialize a property, as shown with the missing `baz`. Another option is to create an assertion function to simulate lazy initialization:

function setProp<T extends object, K extends PropertyKey, V>(
  obj: T, key: K, val: V
): asserts obj is T & { [P in K]: V } {
  (obj as any)[key] = val;
}

The `setProp()` function allows for dynamic property assignments, providing a safer way to extend object types. Here is an example of how to use it in your code:

function BuildBigType(params: string[]): BigType {
  const endResult = {};
  setProp(endResult, "foo", 5);
  setProp(endResult, "bar", undefined);
  setProp(endResult, "baz", 123);
  return endResult; // no issues
}

By using `setProp()`, you ensure proper type checking and extend object types safely. If a property is forgotten, the compiler will raise an error as demonstrated in the `BuildBigTypeError` function.

Check out this Playground link with the code

Answer 4

If you're searching for a way to implement lazy object initialization in TypeScript, you may be interested to know that it is not directly supported. You can check out the open feature request microsoft/TypeScript#28503 and show your support by giving it a thumbs up.

One way to modify your code to avoid errors without resorting to using the potentially unsafe `any` type is to utilize a type assertion when declaring the variable. This would look like:

function BuildBigTypeAssert(params: string[]) {
  const endResult = {} as BigType;
  endResult.foo = 5;
  endResult.bar = undefined;
  return endResult; // no errors, but be cautious
}

While this approach is acceptable, keep in mind that the compiler won't catch an error if you forget to initialize a property, as shown with the missing `baz`. Another option is to create an assertion function to simulate lazy initialization:

function setProp<T extends object, K extends PropertyKey, V>(
  obj: T, key: K, val: V
): asserts obj is T & { [P in K]: V } {
  (obj as any)[key] = val;
}

The `setProp()` function allows for dynamic property assignments, providing a safer way to extend object types. Here is an example of how to use it in your code:

function BuildBigType(params: string[]): BigType {
  const endResult = {};
  setProp(endResult, "foo", 5);
  setProp(endResult, "bar", undefined);
  setProp(endResult, "baz", 123);
  return endResult; // no issues
}

By using `setProp()`, you ensure proper type checking and extend object types safely. If a property is forgotten, the compiler will raise an error as demonstrated in the `BuildBigTypeError` function.

Check out this Playground link with the code

Answer 5

Answer №3

To construct this entity, you can follow a two-step process. Initially, create a mandatory field entity. Subsequently, establish endResult: BigType and input values for optional fields.

export function ConstructBigType(parameters: string[]): BigType {
  // Create mandatory fields first
  const mandatoryFields = { foo: null, baz: null };
  mandatoryFields.foo = 5;
  mandatoryFields.baz = 10;

  const endResult: BigType = { ...mandatoryFields };
 
  // Establish optional fields
  endResult.bar = undefined;

  return endResult;
}

Answer 6

To construct this entity, you can follow a two-step process. Initially, create a mandatory field entity. Subsequently, establish endResult: BigType and input values for optional fields.

export function ConstructBigType(parameters: string[]): BigType {
  // Create mandatory fields first
  const mandatoryFields = { foo: null, baz: null };
  mandatoryFields.foo = 5;
  mandatoryFields.baz = 10;

  const endResult: BigType = { ...mandatoryFields };
 
  // Establish optional fields
  endResult.bar = undefined;

  return endResult;
}

What is a way to construct an object without resorting to casts or manually declaring variables for each field?

Answer №1

Answer №2

Answer №3

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