How to quickly discriminate between interfaces with non-overlapping properties

Question

In typescript, I commonly have a variable with a type like:

const variable: { a: string } | { b: number };

When I want to check what shape the variable has, I usually want to do

if (variable.a !== undefined) {
  // do stuff
}
else {
  // do stuff

However in strict mode, typescript throws a type error at variable.a because variable is not known to have the property a. To get around this, I usually cast to any like (variable as any).a !== undefined. However typescript doesn't treat this as a type guard and doesn't narrow the type of variable.

Is there a "proper" way of doing this check in typescript when in strict mode (without making a custom type guard)?

Some examples I've tried (which, unfortunately, also throw type errors):

• typeof variable.a !== 'undefined'
• variable.hasOwnProperty('a')

Answer 1

The simplest way to discriminate in this case would be an in type guard:

declare const variable: { a: string } | { b: number };


if ('a' in variable) {
    variable.a
}
else {
    variable.b
}