Issue 2339: Dealing with || and Union Types

I've encountered an interesting issue while working with TypeScript and JavaScript. I created a code snippet that runs perfectly in JavaScript but throws a syntax error in TypeScript. You can check it out in action at this TypeScript Sandbox.

Essentially, the problem arises when attempting to use the|| (OR) operator to handle cases where an object property might be undefined. TypeScript raises an error claiming the property doesn't exist, even though I'm using || specifically for that reason.

class Person {
    name: string;
    job: string;
}

var person1:Person = {name: 'hi',   job: 'yes'};
var name = 'bye';

function showName(person1:Person | string ) {
    var personsName: string = person1.name || person1;

    document.write(personsName);
    document.write('<br />');
}

showName(person1);
showName(name);

The expected output is:

hi
bye

However, TypeScript flags a syntax error on person1.name. Why is this happening? And what would be the correct way to address this in TypeScript?

typescripttypescript1.5

Answer №1

One solution I suggest is to implement a type guard:

function displayUsername(user: User | string) {
    var userName = (typeof user === "string") ? user : user.username;

    document.write(userName);
    document.write('<br />');
}

Keep in mind that this method will raise an error if user is null || undefined, just like the original code. There are several approaches to address this issue, so feel free to explore different options.

Answer №2

To handle a union type operand, it is important to explicitly define the type in your expression. For example:

const username: string = (person1 as Person).name || person1 as string;

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