Tips for disentangling code from types in Typescript

Instead of intertwining code and types like the example below:

const compar8 : boolean | error = (action: string, n: number) => {
    switch(action) {
        case 'greater':
            return n > 8;
        case 'less':
            return n < 8;
        case 'equal':
            return n === 8;
        default:
            throw new Error('Invalid action');
    }
}

I prefer to have a setup similar to this:

// code.js

const compar8 = (action, n) => {
    switch(action) {
        case 'greater':
            return n > 8;
        case 'less':
            return n < 8;
        case 'equal':
            return n === 8;
        default:
            throw new Error('Invalid action');
    }
}

// types.ts
compar8 : (string, number) => boolean | error

The motivation behind this preference is twofold: 1. Improved readability and 2. Consistency in assigning specific types to functions. For example:

myType = (string, number) => boolean | error

This way, we can use the same type definition for all functions with identical signatures:

// types.ts

compar8: myType
compar9: myType
...
and so on

Is there a method to achieve this?

Answer №1

To ensure proper handling, define it as a type beforehand:

type CompareFunction = (comparison: string, num: number) => boolean;
const compare: CompareFunction = (comparison, num) => {
    switch(action) {
        case 'greater':
            return num > 8;
        case 'less':
            return num < 8;
        case 'equal':
            return num === 8;
        default:
            throw new Error('Invalid action');
    }
}

It is important to note that if the function does not return an error, the type should not include it. TypeScript does not specify the return type for exceptions that are potentially thrown. You would only use => boolean | error if you were returning an instance of Error(), not using throw.

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