How to do proper type checking?

Question

typescript

There is such a code

const installed = window.getOfflineSigner && window.keplr 

const enable = () => {
  if (installed) return window.keplr.enable(chainId)
}

Keplr is a browser extension so I need to check if it is installed. Since Installed is just bool ts, it doesn't understand that window has keplr. If you pass the full condition, then everything works as it should. But I would like not to write it everywhere. How to make such code work? Naturally without an operator! after keplr.

Answer 1

https://www.typescriptlang.org/docs/handbook/2/nar...

type Keplr = {
  enable(chainId: ChainId): Enable;
};
const isInstalled = (v: unknown): v is Keplr => !!(window.getOfflineSigner && v) 

const enable = () => {
  if (isInstalled(window.keplr)) return window.keplr.enable(chainId)
}

Well, either wait for the release of 4.4 or switch to beta:
https://devblogs.microsoft.com/typescript/announci...

Answer 2

https://www.typescriptlang.org/docs/handbook/relea...