TypeScript - Functionally Speaking

Question

TypeScript - Functionally Speaking

When defining a function type, the example below does not allow the use of const

interface Square{
    readonly color: string;
    readonly area: number;
}

interface compareTo{
    (const v1: Square, const v2: Square): number;  // syntax error      
}

Is there a way to compare v1 and v2 without altering the object that Square is pointing to?

typescript ecmascript-6

Answer 1

Answer №1

Declaring this would be ineffective as const in TypeScript does not signify immutability. It only pertains to bindings, and a function parameter is not an externally-visible binding.

Answer 2

Declaring this would be ineffective as const in TypeScript does not signify immutability. It only pertains to bindings, and a function parameter is not an externally-visible binding.

Answer 3

Answer №2

Explained in another response, the use of const in ES6 does not prevent objects from being modified, but rather only stops reassignments.

To globally prohibit parameter reassignments, one can employ TSLint's no-parameter-reassignment rule.

If the goal is to prevent object modifications during runtime, one should utilize Object.freeze. At compile time, this restriction can be imposed using the Readonly mapped type. However, if the types are compatible, this method will not have any effect:

interface compareTo {
    (v1: Readonly<Square>, v2: Readonly<Square>): number;      
}
const foo: compareTo = (a: Square, b: Square) => {
  a.area = 0;
  return 1;
}

Typically, it is not the role of an interface to dictate how a function functions internally; it simply defines its interface. Therefore, it would be represented as follows:

interface compareTo {
    (v1: Square, v2: Square): number;      
}
const foo: compareTo = (a: Readonly<Square>, b: Readonly<Square>) => {
  a.area = 0; // error
  return 1;
}

This approach will succeed when types are inferred from a generic type, meaning that the type of a is not specified in the implementation of compareTo:

interface compareTo<T = Readonly<Square>> {
    (v1: T, v2: T): number;      
}

const foo: compareTo = (a, b) => {
  a.area = 0; // error
  return 1;
}

Answer 4

Explained in another response, the use of const in ES6 does not prevent objects from being modified, but rather only stops reassignments.

To globally prohibit parameter reassignments, one can employ TSLint's no-parameter-reassignment rule.

If the goal is to prevent object modifications during runtime, one should utilize Object.freeze. At compile time, this restriction can be imposed using the Readonly mapped type. However, if the types are compatible, this method will not have any effect:

interface compareTo {
    (v1: Readonly<Square>, v2: Readonly<Square>): number;      
}
const foo: compareTo = (a: Square, b: Square) => {
  a.area = 0;
  return 1;
}

Typically, it is not the role of an interface to dictate how a function functions internally; it simply defines its interface. Therefore, it would be represented as follows:

interface compareTo {
    (v1: Square, v2: Square): number;      
}
const foo: compareTo = (a: Readonly<Square>, b: Readonly<Square>) => {
  a.area = 0; // error
  return 1;
}

This approach will succeed when types are inferred from a generic type, meaning that the type of a is not specified in the implementation of compareTo:

interface compareTo<T = Readonly<Square>> {
    (v1: T, v2: T): number;      
}

const foo: compareTo = (a, b) => {
  a.area = 0; // error
  return 1;
}

TypeScript - Functionally Speaking

Answer №1

Answer №2

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