In what scenario would one require an 'is' predicate instead of utilizing the 'in' operator?

Question

In what scenario would one require an 'is' predicate instead of utilizing the 'in' operator?

The TypeScript documentation highlights the use of TypeGuards for distinguishing between different types.

Examples in the documentation showcase the is predicate and the in operator for this purpose.

is predicate:

function isFish(pet: Fish | Bird): pet is Fish {
    return (pet as Fish).swim !== undefined;
}

if (isFish(pet)) {
    pet.swim();
}
else {
    pet.fly();
}

in operator:

function move(pet: Fish | Bird) {
    if ("swim" in pet) {
        return pet.swim();
    }
    return pet.fly();
}

The usage of the in operator appears to be more concise, yet extracting type differentiation into a function like the is predicate enhances readability and maintenance in the long term.

Are there any instances where the is predicate can effectively differentiate between types when compared to the in operator?

typescript

Answer 1

Answer №1

The in operator provides a method to narrow the type of a variable. Type guards can be created using various checks, such as:

type JSONPrimitive = number | string | boolean | null;

function isJsonPrimitive(x: any): x is JSONPrimitive {
    return x === null
        || typeof x === 'number'
        || typeof x === 'string'
        || typeof x === 'boolean';
}

Alternatively:

interface TreeNode {
    value: number;
    left: TreeNode | null;
    right: TreeNode | null;
}
type InternalNode = TreeNode & ({ left: TreeNode } | { right: TreeNode });

function isInternalNode(node: TreeNode): node is InternalNode {
    return node.left !== null || node.right !== null;
}

It's worth noting that neither of these examples can be replicated using the in operator.

User-defined type guards are akin to type assertions, with TypeScript not verifying their accuracy. This means you have flexibility in the logic used within your type guard implementation; even a simple return true; is valid.

Answer 2

The in operator provides a method to narrow the type of a variable. Type guards can be created using various checks, such as:

type JSONPrimitive = number | string | boolean | null;

function isJsonPrimitive(x: any): x is JSONPrimitive {
    return x === null
        || typeof x === 'number'
        || typeof x === 'string'
        || typeof x === 'boolean';
}

Alternatively:

interface TreeNode {
    value: number;
    left: TreeNode | null;
    right: TreeNode | null;
}
type InternalNode = TreeNode & ({ left: TreeNode } | { right: TreeNode });

function isInternalNode(node: TreeNode): node is InternalNode {
    return node.left !== null || node.right !== null;
}

It's worth noting that neither of these examples can be replicated using the in operator.

User-defined type guards are akin to type assertions, with TypeScript not verifying their accuracy. This means you have flexibility in the logic used within your type guard implementation; even a simple return true; is valid.

In what scenario would one require an 'is' predicate instead of utilizing the 'in' operator?

Answer №1

Similar questions

Enhancing Typescript Arrow Function Parameters using Decorators

Utilizing CSS files to incorporate loading icons in a component by dynamically updating based on passed props

TypeScript's type assertions do not result in errors when provided with an incorrect value

Angular2 checkboxes for filtering data

Are you facing difficulties while trying to utilize useContext in your React application?

Guidelines for converting an array into checkboxes using React Native with TypeScript

Exploring the synergies between Angular 5 and atmosphere.js

The creation of the Angular project was unsuccessful due to the outdated circular-json dependency

How can I integrate the jQuery Plugin Mapael with Angular 5?

Creating an index signature in TypeScript without having to use union types for the values - a comprehensive guide

Using Typescript to mute audio elements within HTML documents

How can I retrieve the selected value from an Angular 2 dropdown menu when it changes, in order to utilize it within a function?

I am interested in monitoring for any alterations to the @input Object

Creating a TypeScript class with methods to export as an object

Vitest's behavior shows variance when compared to compiled JavaScript

Encounter the "Error: Source 'cloudsTileLayer-RasterSource' not found" message while trying to integrate a weather tile layer into Azure Maps

Ways to ascertain if a TypeScript type is a specific value

Issue: Incompatibility in metadata versions detected for module .../ngx-masonry/ngx-masonry.d.ts. Level 4 version identified, whereas level 3 version

Error in Angular Material: SassError - The CSS after "@include mat" is invalid. Expected 1 selector or at-rule, but found ".core();"

Unable to retrieve this information within an anonymous function