What is the reason behind TypeScript resolving a union type to an empty string in the presence of logical operators?

To prevent a null object error, I attempted to use logical operators. The object in question has attributes including a string and a union type. When I employed the "&&" operator to verify the object's existence, it resulted in evaluating the expression to another union type that includes empty strings, leading to a type mismatch error.

Upon testing with an IF statement instead of the logical operator, the code behaved as expected. However, I am eager to grasp why my original approach did not yield the desired outcome.

type FooType = 'aaa' | 'bbb' | 'ccc';
​
interface IFoo {
  type: FooType;
  id: string;
}
​
type FooAlias = IFoo;
​
const fooObject = {type: 'aaa', id: '111'} as IFoo;
​
const fooObject2: FooAlias = {
  id: fooObject && fooObject.id,
  type: fooObject && fooObject.type
  /* An error is flagged on the line above:
  Type '"" | "aaa" | "bbb" | "ccc"' is not assignable to type 'FooType'.
  Type '""' is not assignable to type 'FooType'.ts(2322)*/
};

Replacing

type: fooObject && fooObject.type
with
type: fooObject ? fooObject.type : null
resolved the issue, suggesting that it appropriately handled scenarios where the value might be an empty string.

Answer №1

This issue has been identified as a bug. When strictNullChecks is disabled, an extra empty string "" may appear.

While waiting for a resolution, you can apply a type assertion or use a ternary operator:

const fooObject2: FooAlias = {
  id: fooObject && fooObject.id,
  //type: (fooObject && fooObject.type) as FooType
  type: fooObject ? fooObject.type : undefined
};

Play

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