TypeScript struggles to determine the type when provided with a combination of function unions

What key element am I overlooking?

type HandlerA = (firstArg: string) => number;
type HandlerB = (firstArg: string, secondArg: number) => string;

type Handler = HandlerA | HandlerB

// success
const testA: Handler = (a, b) => 'text'

// failure, as a is assumed to be of type 'any'
const testB: Handler = (a) => 10

Upon analysis, it becomes evident that testB cannot infer HandlerA. Is there an alternative approach to rectify this without explicitly defining like 

const testC: Handler = (a: string) => 10

typescript

Answer №1

When utilizing a union of function types in TypeScript, the type inference may fail if the provided function does not align with all signatures. For example, in testB, the failure occurs because HandlerB expects two arguments but only one is given, resulting in an error. Check out this issue highlighted by @jcalz

Are you attempting to perform function overloading?

function handler(firstArg: string): string
function handler(firstArg: string, secondArg: number): string 
function handler(firstArg: string, secondArg?: number): string | number {
  if (!!firstArg && !secondArg) {
   return   'implementation a'
  }
  return 'implementation b'
}

Learn more about TypeScript function overloading here

You could also define a more generic handler type:

type Handler = (firstArg: string, secondArg?: number) => string | number;

const testA: Handler = (a, b) => 'text'
const testB: Handler = (a) => 10

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