[Solved]-React prop types with TypeScript - how to have a function type?-Reactjs

Concerning Function and best practices

Function in practice has unsufficient type-safety, being the super-type of all functions. You better replace it with a function type - see the "Solutions" section down under.

In above example, Function is not assignable to the more narrow onClick function type, which causes the error at hand (Playground example).

In addition to mentioned issue, here is what TypeScript docs say about Function:

This is an untyped function call and is generally best avoided because of the unsafe any return type. If need to accept an arbitrary function but don’t intend to call it, the type () => void is generally safer.

typescript-eslint has discarded Function with the ban-types rule, emitting following message with default configuration (see also here):

The Function type accepts any function-like value. It provides no type safety when calling the function, which can be a common source of bugs. If you are expecting the function to accept certain arguments, you should explicitly define the function shape.

Better solutions

React already comes with built-in event handler-types to handle common events.

type Props = {
  onClick: React.MouseEventHandler<HTMLButtonElement>
};

const Submit = ({ onClick }: Props) => <button onClick={onClick}> click </button>

type Props = { 
  onClick: (event: React.MouseEvent<HTMLElement>) => void
};

void is more restrictive than any. There is no chance to return a value in the callback by accident, which would be possible with any.

In summary, we embrace typing capabilities of TypeScript instead of using Function or any. The parameter now is known to be MouseEvent and the return type void, identifying it as a callback.

Related

Typescript: How to define type for a function callback (as any function type, not universal any) used in a method parameter