TypeScript - void function passes as a a plain object, why?

Question

Why does a void function pass as a plain object? Is there a good reason for this, or is it a bug?

type PlainObject = { [key: string]: any };

const foo = (value: PlainObject) => { }

const voidFn: () => void = () => { };

// Error as expected
foo(false);
foo(3);
foo('test');

foo ({ test: true }) // ok - expected usage

foo(voidFn); // Why does it not trigger an error?

The problem occurs only if the value is any. If it was { [key: string]: string }; for instance, the error would occur.

Almost everything in JavaScript is an object. In fact, only six things are not objects. They are — null , undefined , strings, numbers, boolean, and symbols. These are called primitive values or primitive types — Ernesto
– Ernesto, Commented Apr 10, 2021 at 16:03

Quentin · Accepted Answer · 2021-04-10 14:51:36Z

2

Functions are objects.

const f = () => null;
const o = {};

console.log(f instanceof Object);
console.log(o instanceof Object);

Since every property of the PlainObject can have any value, it doesn't matter what properties the function has.

answered Apr 10, 2021 at 14:51

Quentin

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1 Comment

Maciej Kravchyk Over a year ago

Makes sense. For some reason, I thought TypeScript would abstract that away in the types, but I see how even a (() => { }).name is a valid operation, so it has to be that way.

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