strictFunctionTypes

Parameters compare the other way round

A function that accepts less cannot stand in for one that must accept more — obvious when stated, and unchecked without this flag.

Since: TypeScript 2.6
strict: In strict
In your tsconfig: "strictFunctionTypes": true
Compiled with: "strict": true

The same snippet both times. Only the option changed.

With it off

"strictFunctionTypes": false

type Handler = (value: string | number) => void

declare const narrow: (value: string) => void

export const handler: Handler = narrow

Compiles clean

With it on

"strictFunctionTypes": true

type Handler = (value: string | number) => void

declare const narrow: (value: string) => void

export const handler: Handler = narrow

Emits TS2322

Why the compiler bothers

Return types compare covariantly, which everyone expects. Parameters compare contravariantly, which nobody does until it is pointed out: the replacement has to handle every input the original promised to. TypeScript checked parameters bivariantly for years because real code — especially the DOM — relies on it, and this flag turns on the sound rule for function types while deliberately leaving method parameters bivariant, which is why `interface A { f(x: string): void }` is still accepted where the arrow form is not.

Takeaway

Write callbacks as arrow-typed properties, not methods, if you want them checked.

Where to go next

22 options