in operator

Learn narrowing by property with the in operator.

in operator

The in operator checks whether a property exists on an object. A check like "swim" in animal lets TypeScript narrow a union type; if the property exists only on one member, the object is narrowed to that member's type inside the block.

interface Fish { swim(): void }
interface Bird { fly(): void }

function move(animal: Fish | Bird) {
  if ("swim" in animal) {
    animal.swim();  // Fish
  } else {
    animal.fly();  // Bird
  }
}

Distinguishing object shapes

The in operator is ideal for distinguishing objects that have different fields rather than a common one. If each member has a unique property, checking it with in narrows the type. Discriminated unions use a common kind field, but in is more flexible.

interface Square { size: number }
interface Circle { radius: number }

function area(shape: Square | Circle): number {
  if ("radius" in shape) {
    return Math.PI * shape.radius ** 2;  // Circle
  }
  return shape.size ** 2;  // Square
}

Exercise Separate shapes with in

Define Square (size: number) and Circle (radius: number) interfaces. Write an area(shape: Square | Circle): number function. If "radius" in shape, return Math.PI * shape.radius ** 2; otherwise return shape.size ** 2. Print results via console.log(area({ size: 5 })) and console.log(area({ radius: 2 })).

Show example solution

interface Square { size: number; }
interface Circle { radius: number; }

function area(shape: Square | Circle): number {
  if ("radius" in shape) {
    return Math.PI * shape.radius ** 2;
  }
  return shape.size ** 2;
}

console.log(area({ size: 5 }));
console.log(area({ radius: 2 }));

Test yourself

What does the in operator check?
- The type of a value
- The superclass of a class
- Whether a property exists on an object
- Whether two objects have the same shape
Show answer

Whether a property exists on an object — The in operator returns whether the property on the left exists on the object on the right; this is used to distinguish members of a union that have different shapes.

Property check with in

interface Fish { swim(): void }
interface Bird { fly(): void }

function move(a: Fish | Bird) {
  if (____) {
    a.swim();
  } else {
    a.fly();
  }
}

Show answer

interface Fish { swim(): void }
interface Bird { fly(): void }

function move(a: Fish | Bird) {
  if ("swim" in a) {
    a.swim();
  } else {
    a.fly();
  }
}

in operator parts

in operatörü

property exists on object
"swim" in animal

narrows to Fish member
interface

defines object shape
Discriminated union

distinguish via common kind field

Next lesson: instanceof

We Are Enhancing Your Experience

Cookie Preferences

Strictly Necessary Cookies

Analytical Cookies