Type validation in TypeScript through cross-referencing variables

Can TypeScript be used to define a variable that determines the type of other variables? I want to simplify the process by only needing to check one variable, stateIndicator, which is dependent on other variables to infer their types.

type A = {prop1: string}
type B = {prop2: number}

function isA(x: A | B): x is A {
  return (x as A).prop1 !== undefined
}

function foo1(data: A | B) {
  const stateIndicator = isA(data);
  if (stateIndicator) {
    // this is fine
    const myString: string = data.prop1;
  }
}

function foo2(data: A | B) {
  const stateIndicator = isA(data) ? "stateA" : "stateB";
  if (stateIndicator === "stateA") {
    // without checking if isA(data) or using 'data as A', data.prop1 will show error 
    // Property 'prop1' does not exist on type 'B'
    const myString: string = data.prop1;
  }
}

Answer №1

Sorry, but the functionality you're seeking cannot be achieved in TypeScript. Please refer to microsoft/TypeScript#46915 for a detailed explanation.


In essence, the TypeScript compiler is unable to predict all potential logical outcomes of your code. It does not engage in counterfactual reasoning where it makes assumptions based on hypothetical scenarios within the code. This level of analysis would be too complex and resource-intensive to implement universally.

In most cases, when you perform checks on variables, they will only affect the type inference of that specific variable and not others linked to it. Even when checking object properties, the impact is limited to the property's type without altering the overall type of the parent object, unless working with discriminated unions.

There are exceptions to this behavior, such as the control flow analysis introduced in TypeScript 4.4 which allows narrowing based on aliased conditions and discriminants. But for your specific scenario in `foo2`, no such narrowing takes place.


Given the limitations, you may need to find a workaround. One common approach is transforming the union into a discriminated union by adding a discriminant property to each member and then performing checks based on that property:

type A = { stateIndicator: "stateA", prop1: string }
type B = { stateIndicator: "stateB", prop2: number }

function foo(data: A | B) {
    if (data.stateIndicator === "stateA") {
        const myString: string = data.prop1;
    } else {
        const myNumber: number = data.prop2;
    }
}

Your specific use case will dictate the best workaround strategy to adopt.

Link to Playground with sample code

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