Advanced Types

This page lists some of the more advanced ways in which you can model types, it works in tandem with the Utility Types doc which includes types which are included in TypeScript and available globally.

Type Guards and Differentiating Types

Union types are useful for modeling situations when values can overlap in the types they can take on. What happens when we need to know specifically whether we have a Fish? A common idiom in JavaScript to differentiate between two possible values is to check for the presence of a member. As we mentioned, you can only access members that are guaranteed to be in all the constituents of a union type.

ts
let pet = getSmallPet(); // Each of these property accesses will cause an error if (pet.swim) { pet.swim(); } else if (pet.fly) { pet.fly(); }

To get the same code working, we’ll need to use a type assertion:

ts
let pet = getSmallPet(); if ((pet as Fish).swim) { (pet as Fish).swim(); } else if ((pet as Bird).fly) { (pet as Bird).fly(); }

User-Defined Type Guards

Notice that we had to use type assertions several times. It would be much better if once we performed the check, we could know the type of pet within each branch.

It just so happens that TypeScript has something called a type guard. A type guard is some expression that performs a runtime check that guarantees the type in some scope.

Using type predicates

To define a type guard, we simply need to define a function whose return type is a type predicate:

ts
function isFish(pet: Fish | Bird): pet is Fish { return (pet as Fish).swim !== undefined; }

pet is Fish is our type predicate in this example. A predicate takes the form parameterName is Type, where parameterName must be the name of a parameter from the current function signature.

Any time isFish is called with some variable, TypeScript will narrow that variable to that specific type if the original type is compatible.

ts
// Both calls to 'swim' and 'fly' are now okay. if (isFish(pet)) { pet.swim(); } else { pet.fly(); }

Notice that TypeScript not only knows that pet is a Fish in the if branch; it also knows that in the else branch, you don’t have a Fish, so you must have a Bird.

Using the in operator

The in operator now acts as a narrowing expression for types.

For a n in x expression, where n is a string literal or string literal type and x is a union type, the “true” branch narrows to types which have an optional or required property n, and the “false” branch narrows to types which have an optional or missing property n.

ts
function move(pet: Fish | Bird) { if ("swim" in pet) { return pet.swim(); } return pet.fly(); }

typeof type guards

Let’s go back and write the code for the version of padLeft that uses union types. We could write it with type predicates as follows:

ts
function isNumber(x: any): x is number { return typeof x === "number"; } function isString(x: any): x is string { return typeof x === "string"; } function padLeft(value: string, padding: string | number) { if (isNumber(padding)) { return Array(padding + 1).join(" ") + value; } if (isString(padding)) { return padding + value; } throw new Error(`Expected string or number, got '${padding}'.`); }

However, having to define a function to figure out if a type is a primitive is kind of a pain. Luckily, you don’t need to abstract typeof x === "number" into its own function because TypeScript will recognize it as a type guard on its own. That means we could just write these checks inline.

ts
function padLeft(value: string, padding: string | number) { if (typeof padding === "number") { return Array(padding + 1).join(" ") + value; } if (typeof padding === "string") { return padding + value; } throw new Error(`Expected string or number, got '${padding}'.`); }

These typeof type guards are recognized in two different forms: typeof v === "typename" and typeof v !== "typename", where "typename" must be "number", "string", "boolean", or "symbol". While TypeScript won’t stop you from comparing to other strings, the language won’t recognize those expressions as type guards.

instanceof type guards

If you’ve read about typeof type guards and are familiar with the instanceof operator in JavaScript, you probably have some idea of what this section is about.

instanceof type guards are a way of narrowing types using their constructor function. For instance, let’s borrow our industrial string-padder example from earlier:

ts
interface Padder { getPaddingString(): string; } class SpaceRepeatingPadder implements Padder { constructor(private numSpaces: number) {} getPaddingString() { return Array(this.numSpaces + 1).join(" "); } } class StringPadder implements Padder { constructor(private value: string) {} getPaddingString() { return this.value; } } function getRandomPadder() { return Math.random() < 0.5 ? new SpaceRepeatingPadder(4) : new StringPadder(" "); } // Type is 'SpaceRepeatingPadder | StringPadder' let padder: Padder = getRandomPadder(); if (padder instanceof SpaceRepeatingPadder) { padder; // type narrowed to 'SpaceRepeatingPadder' } if (padder instanceof StringPadder) { padder; // type narrowed to 'StringPadder' }

The right side of the instanceof needs to be a constructor function, and TypeScript will narrow down to:

  1. the type of the function’s prototype property if its type is not any
  2. the union of types returned by that type’s construct signatures

in that order.

Nullable types

TypeScript has two special types, null and undefined, that have the values null and undefined respectively. We mentioned these briefly in the Basic Types section. By default, the type checker considers null and undefined assignable to anything. Effectively, null and undefined are valid values of every type. That means it’s not possible to stop them from being assigned to any type, even when you would like to prevent it. The inventor of null, Tony Hoare, calls this his “billion dollar mistake”.

The --strictNullChecks flag fixes this: when you declare a variable, it doesn’t automatically include null or undefined. You can include them explicitly using a union type:

ts
let s = "foo"; s = null; // error, 'null' is not assignable to 'string' let sn: string | null = "bar"; sn = null; // ok sn = undefined; // error, 'undefined' is not assignable to 'string | null'

Note that TypeScript treats null and undefined differently in order to match JavaScript semantics. string | null is a different type than string | undefined and string | undefined | null.

From TypeScript 3.7 and onwards, you can use optional chaining to simplify working with nullable types.

Optional parameters and properties

With --strictNullChecks, an optional parameter automatically adds | undefined:

ts
function f(x: number, y?: number) { return x + (y || 0); } f(1, 2); f(1); f(1, undefined); f(1, null); // error, 'null' is not assignable to 'number | undefined'

The same is true for optional properties:

ts
class C { a: number; b?: number; } let c = new C(); c.a = 12; c.a = undefined; // error, 'undefined' is not assignable to 'number' c.b = 13; c.b = undefined; // ok c.b = null; // error, 'null' is not assignable to 'number | undefined'

Type guards and type assertions

Since nullable types are implemented with a union, you need to use a type guard to get rid of the null. Fortunately, this is the same code you’d write in JavaScript:

ts
function f(sn: string | null): string { if (sn == null) { return "default"; } else { return sn; } }

The null elimination is pretty obvious here, but you can use terser operators too:

ts
function f(sn: string | null): string { return sn || "default"; }

In cases where the compiler can’t eliminate null or undefined, you can use the type assertion operator to manually remove them. The syntax is postfix !: identifier! removes null and undefined from the type of identifier:

ts
function broken(name: string | null): string { function postfix(epithet: string) { return name.charAt(0) + ". the " + epithet; // error, 'name' is possibly null } name = name || "Bob"; return postfix("great"); } function fixed(name: string | null): string { function postfix(epithet: string) { return name!.charAt(0) + ". the " + epithet; // ok } name = name || "Bob"; return postfix("great"); }

The example uses a nested function here because the compiler can’t eliminate nulls inside a nested function (except immediately-invoked function expressions). That’s because it can’t track all calls to the nested function, especially if you return it from the outer function. Without knowing where the function is called, it can’t know what the type of name will be at the time the body executes.

Type Aliases

Type aliases create a new name for a type. Type aliases are sometimes similar to interfaces, but can name primitives, unions, tuples, and any other types that you’d otherwise have to write by hand.

ts
type Name = string; type NameResolver = () => string; type NameOrResolver = Name | NameResolver; function getName(n: NameOrResolver): Name { if (typeof n === "string") { return n; } else { return n(); } }

Aliasing doesn’t actually create a new type - it creates a new name to refer to that type. Aliasing a primitive is not terribly useful, though it can be used as a form of documentation.

Just like interfaces, type aliases can also be generic - we can just add type parameters and use them on the right side of the alias declaration:

ts
type Container<T> = { value: T };

We can also have a type alias refer to itself in a property:

ts
type Tree<T> = { value: T; left: Tree<T>; right: Tree<T>; };

Together with intersection types, we can make some pretty mind-bending types:

ts
type LinkedList<T> = T & { next: LinkedList<T> }; interface Person { name: string; } var people: LinkedList<Person>; var s = people.name; var s = people.next.name; var s = people.next.next.name; var s = people.next.next.next.name;

However, it’s not possible for a type alias to appear anywhere else on the right side of the declaration:

ts
type Yikes = Array<Yikes>; // error

Interfaces vs. Type Aliases

As we mentioned, type aliases can act sort of like interfaces; however, there are some subtle differences.

One difference is that interfaces create a new name that is used everywhere. Type aliases don’t create a new name — for instance, error messages won’t use the alias name. In the code below, hovering over interfaced in an editor will show that it returns an Interface, but will show that aliased returns object literal type.

ts
type Alias = { num: number }; interface Interface { num: number; } declare function aliased(arg: Alias): Alias; declare function interfaced(arg: Interface): Interface;

In older versions of TypeScript, type aliases couldn’t be extended or implemented from (nor could they extend/implement other types). As of version 2.7, type aliases can be extended by creating a new intersection type e.g. type Cat = Animal & { purrs: true }.

Because an ideal property of software is being open to extension, you should always use an interface over a type alias if possible.

On the other hand, if you can’t express some shape with an interface and you need to use a union or tuple type, type aliases are usually the way to go.

String Literal Types

String literal types allow you to specify the exact value a string must have. In practice string literal types combine nicely with union types, type guards, and type aliases. You can use these features together to get enum-like behavior with strings.

ts
type Easing = "ease-in" | "ease-out" | "ease-in-out"; class UIElement { animate(dx: number, dy: number, easing: Easing) { if (easing === "ease-in") { // ... } else if (easing === "ease-out") { } else if (easing === "ease-in-out") { } else { // error! should not pass null or undefined. } } } let button = new UIElement(); button.animate(0, 0, "ease-in"); button.animate(0, 0, "uneasy"); // error: "uneasy" is not allowed here

You can pass any of the three allowed strings, but any other string will give the error

Argument of type '"uneasy"' is not assignable to parameter of type '"ease-in" | "ease-out" | "ease-in-out"'

String literal types can be used in the same way to distinguish overloads:

ts
function createElement(tagName: "img"): HTMLImageElement; function createElement(tagName: "input"): HTMLInputElement; // ... more overloads ... function createElement(tagName: string): Element { // ... code goes here ... }

Numeric Literal Types

TypeScript also has numeric literal types.

ts
function rollDice(): 1 | 2 | 3 | 4 | 5 | 6 { // ... }

These are seldom written explicitly, but they can be useful when narrowing issues and can catch bugs:

ts
function foo(x: number) { if (x !== 1 || x !== 2) { // ~~~~~~~ // Operator '!==' cannot be applied to types '1' and '2'. } }

In other words, x must be 1 when it gets compared to 2, meaning that the above check is making an invalid comparison.

Enum Member Types

As mentioned in our section on enums, enum members have types when every member is literal-initialized.

Much of the time when we talk about “singleton types”, we’re referring to both enum member types as well as numeric/string literal types, though many users will use “singleton types” and “literal types” interchangeably.

Discriminated Unions

You can combine singleton types, union types, type guards, and type aliases to build an advanced pattern called discriminated unions, also known as tagged unions or algebraic data types. Discriminated unions are useful in functional programming. Some languages automatically discriminate unions for you; TypeScript instead builds on JavaScript patterns as they exist today. There are three ingredients:

  1. Types that have a common, singleton type property — the discriminant.
  2. A type alias that takes the union of those types — the union.
  3. Type guards on the common property.
ts
interface Square { kind: "square"; size: number; } interface Rectangle { kind: "rectangle"; width: number; height: number; } interface Circle { kind: "circle"; radius: number; }

First we declare the interfaces we will union. Each interface has a kind property with a different string literal type. The kind property is called the discriminant or tag. The other properties are specific to each interface. Notice that the interfaces are currently unrelated. Let’s put them into a union:

ts
type Shape = Square | Rectangle | Circle;

Now let’s use the discriminated union:

ts
function area(s: Shape) { switch (s.kind) { case "square": return s.size * s.size; case "rectangle": return s.height * s.width; case "circle": return Math.PI * s.radius ** 2; } }

Exhaustiveness checking

We would like the compiler to tell us when we don’t cover all variants of the discriminated union. For example, if we add Triangle to Shape, we need to update area as well:

ts
type Shape = Square | Rectangle | Circle | Triangle; function area(s: Shape) { switch (s.kind) { case "square": return s.size * s.size; case "rectangle": return s.height * s.width; case "circle": return Math.PI * s.radius ** 2; } // should error here - we didn't handle case "triangle" }

There are two ways to do this. The first is to turn on --strictNullChecks and specify a return type:

ts
function area(s: Shape): number { // error: returns number | undefined switch (s.kind) { case "square": return s.size * s.size; case "rectangle": return s.height * s.width; case "circle": return Math.PI * s.radius ** 2; } }

Because the switch is no longer exhaustive, TypeScript is aware that the function could sometimes return undefined. If you have an explicit return type number, then you will get an error that the return type is actually number | undefined. However, this method is quite subtle and, besides, --strictNullChecks does not always work with old code.

The second method uses the never type that the compiler uses to check for exhaustiveness:

ts
function assertNever(x: never): never { throw new Error("Unexpected object: " + x); } function area(s: Shape) { switch (s.kind) { case "square": return s.size * s.size; case "rectangle": return s.height * s.width; case "circle": return Math.PI * s.radius ** 2; default: return assertNever(s); // error here if there are missing cases } }

Here, assertNever checks that s is of type never — the type that’s left after all other cases have been removed. If you forget a case, then s will have a real type and you will get a type error. This method requires you to define an extra function, but it’s much more obvious when you forget it.

Polymorphic this types

A polymorphic this type represents a type that is the subtype of the containing class or interface. This is called F-bounded polymorphism. This makes hierarchical fluent interfaces much easier to express, for example. Take a simple calculator that returns this after each operation:

ts
class BasicCalculator { public constructor(protected value: number = 0) {} public currentValue(): number { return this.value; } public add(operand: number): this { this.value += operand; return this; } public multiply(operand: number): this { this.value *= operand; return this; } // ... other operations go here ... } let v = new BasicCalculator(2) .multiply(5) .add(1) .currentValue();

Since the class uses this types, you can extend it and the new class can use the old methods with no changes.

ts
class ScientificCalculator extends BasicCalculator { public constructor(value = 0) { super(value); } public sin() { this.value = Math.sin(this.value); return this; } // ... other operations go here ... } let v = new ScientificCalculator(2) .multiply(5) .sin() .add(1) .currentValue();

Without this types, ScientificCalculator would not have been able to extend BasicCalculator and keep the fluent interface. multiply would have returned BasicCalculator, which doesn’t have the sin method. However, with this types, multiply returns this, which is ScientificCalculator here.

Index types

With index types, you can get the compiler to check code that uses dynamic property names. For example, a common JavaScript pattern is to pick a subset of properties from an object:

js
function pluck(o, propertyNames) { return propertyNames.map(n => o[n]); }

Here’s how you would write and use this function in TypeScript, using the index type query and indexed access operators:

ts
function pluck<T, K extends keyof T>(o: T, propertyNames: K[]): T[K][] { return propertyNames.map(n => o[n]); } interface Car { manufacturer: string; model: string; year: number; } let taxi: Car = { manufacturer: "Toyota", model: "Camry", year: 2014 }; // Manufacturer and model are both of type string, // so we can pluck them both into a typed string array let makeAndModel: string[] = pluck(taxi, ["manufacturer", "model"]); // If we try to pluck model and year, we get an // array of a union type: (string | number)[] let modelYear = pluck(taxi, ["model", "year"]);

The compiler checks that manufacturer and model are actually properties on Car. The example introduces a couple of new type operators. First is keyof T, the index type query operator. For any type T, keyof T is the union of known, public property names of T. For example:

ts
let carProps: keyof Car; // the union of ("manufacturer" | "model" | "year")

keyof Car is completely interchangeable with "manufacturer" | "model" | "year". The difference is that if you add another property to Car, say ownersAddress: string, then keyof Car will automatically update to be "manufacturer" | "model" | "year" | "ownersAddress". And you can use keyof in generic contexts like pluck, where you can’t possibly know the property names ahead of time. That means the compiler will check that you pass the right set of property names to pluck:

ts
// error, Type '"unknown"' is not assignable to type '"manufacturer" | "model" | "year"' pluck(taxi, ["year", "unknown"]);

The second operator is T[K], the indexed access operator. Here, the type syntax reflects the expression syntax. That means that person["name"] has the type Person["name"] — which in our example is just string. However, just like index type queries, you can use T[K] in a generic context, which is where its real power comes to life. You just have to make sure that the type variable K extends keyof T. Here’s another example with a function named getProperty.

ts
function getProperty<T, K extends keyof T>(o: T, propertyName: K): T[K] { return o[propertyName]; // o[propertyName] is of type T[K] }

In getProperty, o: T and propertyName: K, so that means o[propertyName]: T[K]. Once you return the T[K] result, the compiler will instantiate the actual type of the key, so the return type of getProperty will vary according to which property you request.

ts
let name: string = getProperty(taxi, "manufacturer"); let year: number = getProperty(taxi, "year"); // error, Argument of type '"unknown"' is not assignable to parameter of type '"manufacturer" | "model" | "year"' let unknown = getProperty(taxi, "unknown");

Index types and index signatures

keyof and T[K] interact with index signatures. An index signature parameter type must be ‘string’ or ‘number’. If you have a type with a string index signature, keyof T will be string | number (and not just string, since in JavaScript you can access an object property either by using strings (object["42"]) or numbers (object[42])). And T[string] is just the type of the index signature:

ts
interface Dictionary<T> { [key: string]: T; } let keys: keyof Dictionary<number>; // string | number let value: Dictionary<number>["foo"]; // number

If you have a type with a number index signature, keyof T will just be number.

ts
interface Dictionary<T> { [key: number]: T; } let keys: keyof Dictionary<number>; // number let value: Dictionary<number>["foo"]; // Error, Property 'foo' does not exist on type 'Dictionary<number>'. let value: Dictionary<number>[42]; // number

Mapped types

A common task is to take an existing type and make each of its properties optional:

ts
interface PersonPartial { name?: string; age?: number; }

Or we might want a readonly version:

ts
interface PersonReadonly { readonly name: string; readonly age: number; }

This happens often enough in JavaScript that TypeScript provides a way to create new types based on old types — mapped types. In a mapped type, the new type transforms each property in the old type in the same way. For example, you can make all properties of a type readonly or optional. Here are a couple of examples:

ts
type Readonly<T> = { readonly [P in keyof T]: T[P]; }; type Partial<T> = { [P in keyof T]?: T[P]; };

And to use it:

ts
type PersonPartial = Partial<Person>; type ReadonlyPerson = Readonly<Person>;

Note that this syntax describes a type rather than a member. If you want to add members, you can use an intersection type:

ts
// Use this: type PartialWithNewMember<T> = { [P in keyof T]?: T[P]; } & { newMember: boolean } // **Do not** use the following! // This is an error! type PartialWithNewMember<T> = { [P in keyof T]?: T[P]; newMember: boolean; }

Let’s take a look at the simplest mapped type and its parts:

ts
type Keys = "option1" | "option2"; type Flags = { [K in Keys]: boolean };

The syntax resembles the syntax for index signatures with a for .. in inside. There are three parts:

  1. The type variable K, which gets bound to each property in turn.
  2. The string literal union Keys, which contains the names of properties to iterate over.
  3. The resulting type of the property.

In this simple example, Keys is a hard-coded list of property names and the property type is always boolean, so this mapped type is equivalent to writing:

ts
type Flags = { option1: boolean; option2: boolean; };

Real applications, however, look like Readonly or Partial above. They’re based on some existing type, and they transform the properties in some way. That’s where keyof and indexed access types come in:

ts
type NullablePerson = { [P in keyof Person]: Person[P] | null }; type PartialPerson = { [P in keyof Person]?: Person[P] };

But it’s more useful to have a general version.

ts
type Nullable<T> = { [P in keyof T]: T[P] | null }; type Partial<T> = { [P in keyof T]?: T[P] };

In these examples, the properties list is keyof T and the resulting type is some variant of T[P]. This is a good template for any general use of mapped types. That’s because this kind of transformation is homomorphic, which means that the mapping applies only to properties of T and no others. The compiler knows that it can copy all the existing property modifiers before adding any new ones. For example, if Person.name was readonly, Partial<Person>.name would be readonly and optional.

Here’s one more example, in which T[P] is wrapped in a Proxy<T> class:

ts
type Proxy<T> = { get(): T; set(value: T): void; }; type Proxify<T> = { [P in keyof T]: Proxy<T[P]>; }; function proxify<T>(o: T): Proxify<T> { // ... wrap proxies ... } let proxyProps = proxify(props);

Note that Readonly<T> and Partial<T> are so useful, they are included in TypeScript’s standard library along with Pick and Record:

ts
type Pick<T, K extends keyof T> = { [P in K]: T[P]; }; type Record<K extends keyof any, T> = { [P in K]: T; };

Readonly, Partial and Pick are homomorphic whereas Record is not. One clue that Record is not homomorphic is that it doesn’t take an input type to copy properties from:

ts
type ThreeStringProps = Record<"prop1" | "prop2" | "prop3", string>;

Non-homomorphic types are essentially creating new properties, so they can’t copy property modifiers from anywhere.

Inference from mapped types

Now that you know how to wrap the properties of a type, the next thing you’ll want to do is unwrap them. Fortunately, that’s pretty easy:

ts
function unproxify<T>(t: Proxify<T>): T { let result = {} as T; for (const k in t) { result[k] = t[k].get(); } return result; } let originalProps = unproxify(proxyProps);

Note that this unwrapping inference only works on homomorphic mapped types. If the mapped type is not homomorphic you’ll have to give an explicit type parameter to your unwrapping function.

Conditional Types

TypeScript 2.8 introduces conditional types which add the ability to express non-uniform type mappings. A conditional type selects one of two possible types based on a condition expressed as a type relationship test:

ts
T extends U ? X : Y

The type above means when T is assignable to U the type is X, otherwise the type is Y.

A conditional type T extends U ? X : Y is either resolved to X or Y, or deferred because the condition depends on one or more type variables. When T or U contains type variables, whether to resolve to X or Y, or to defer, is determined by whether or not the type system has enough information to conclude that T is always assignable to U.

As an example of some types that are immediately resolved, we can take a look at the following example:

ts
declare function f<T extends boolean>(x: T): T extends true ? string : number; // Type is 'string | number' let x = f(Math.random() < 0.5);

Another example would be the TypeName type alias, which uses nested conditional types:

ts
type TypeName<T> = T extends string ? "string" : T extends number ? "number" : T extends boolean ? "boolean" : T extends undefined ? "undefined" : T extends Function ? "function" : "object"; type T0 = TypeName<string>; // "string" type T1 = TypeName<"a">; // "string" type T2 = TypeName<true>; // "boolean" type T3 = TypeName<() => void>; // "function" type T4 = TypeName<string[]>; // "object"

But as an example of a place where conditional types are deferred - where they stick around instead of picking a branch - would be in the following:

ts
interface Foo { propA: boolean; propB: boolean; } declare function f<T>(x: T): T extends Foo ? string : number; function foo<U>(x: U) { // Has type 'U extends Foo ? string : number' let a = f(x); // This assignment is allowed though! let b: string | number = a; }

In the above, the variable a has a conditional type that hasn’t yet chosen a branch. When another piece of code ends up calling foo, it will substitute in U with some other type, and TypeScript will re-evaluate the conditional type, deciding whether it can actually pick a branch.

In the meantime, we can assign a conditional type to any other target type as long as each branch of the conditional is assignable to that target. So in our example above we were able to assign U extends Foo ? string : number to string | number since no matter what the conditional evaluates to, it’s known to be either string or number.

Distributive conditional types

Conditional types in which the checked type is a naked type parameter are called distributive conditional types. Distributive conditional types are automatically distributed over union types during instantiation. For example, an instantiation of T extends U ? X : Y with the type argument A | B | C for T is resolved as (A extends U ? X : Y) | (B extends U ? X : Y) | (C extends U ? X : Y).

Example

ts
type T10 = TypeName<string | (() => void)>; // "string" | "function" type T12 = TypeName<string | string[] | undefined>; // "string" | "object" | "undefined" type T11 = TypeName<string[] | number[]>; // "object"

In instantiations of a distributive conditional type T extends U ? X : Y, references to T within the conditional type are resolved to individual constituents of the union type (i.e. T refers to the individual constituents after the conditional type is distributed over the union type). Furthermore, references to T within X have an additional type parameter constraint U (i.e. T is considered assignable to U within X).

Example

ts
type BoxedValue<T> = { value: T }; type BoxedArray<T> = { array: T[] }; type Boxed<T> = T extends any[] ? BoxedArray<T[number]> : BoxedValue<T>; type T20 = Boxed<string>; // BoxedValue<string>; type T21 = Boxed<number[]>; // BoxedArray<number>; type T22 = Boxed<string | number[]>; // BoxedValue<string> | BoxedArray<number>;

Notice that T has the additional constraint any[] within the true branch of Boxed<T> and it is therefore possible to refer to the element type of the array as T[number]. Also, notice how the conditional type is distributed over the union type in the last example.

The distributive property of conditional types can conveniently be used to filter union types:

ts
type Diff<T, U> = T extends U ? never : T; // Remove types from T that are assignable to U type Filter<T, U> = T extends U ? T : never; // Remove types from T that are not assignable to U type T30 = Diff<"a" | "b" | "c" | "d", "a" | "c" | "f">; // "b" | "d" type T31 = Filter<"a" | "b" | "c" | "d", "a" | "c" | "f">; // "a" | "c" type T32 = Diff<string | number | (() => void), Function>; // string | number type T33 = Filter<string | number | (() => void), Function>; // () => void type NonNullable<T> = Diff<T, null | undefined>; // Remove null and undefined from T type T34 = NonNullable<string | number | undefined>; // string | number type T35 = NonNullable<string | string[] | null | undefined>; // string | string[] function f1<T>(x: T, y: NonNullable<T>) { x = y; // Ok y = x; // Error } function f2<T extends string | undefined>(x: T, y: NonNullable<T>) { x = y; // Ok y = x; // Error let s1: string = x; // Error let s2: string = y; // Ok }

Conditional types are particularly useful when combined with mapped types:

ts
type FunctionPropertyNames<T> = { [K in keyof T]: T[K] extends Function ? K : never; }[keyof T]; type FunctionProperties<T> = Pick<T, FunctionPropertyNames<T>>; type NonFunctionPropertyNames<T> = { [K in keyof T]: T[K] extends Function ? never : K; }[keyof T]; type NonFunctionProperties<T> = Pick<T, NonFunctionPropertyNames<T>>; interface Part { id: number; name: string; subparts: Part[]; updatePart(newName: string): void; } type T40 = FunctionPropertyNames<Part>; // "updatePart" type T41 = NonFunctionPropertyNames<Part>; // "id" | "name" | "subparts" type T42 = FunctionProperties<Part>; // { updatePart(newName: string): void } type T43 = NonFunctionProperties<Part>; // { id: number, name: string, subparts: Part[] }

Similar to union and intersection types, conditional types are not permitted to reference themselves recursively. For example the following is an error.

Example

ts
type ElementType<T> = T extends any[] ? ElementType<T[number]> : T; // Error

Type inference in conditional types

Within the extends clause of a conditional type, it is now possible to have infer declarations that introduce a type variable to be inferred. Such inferred type variables may be referenced in the true branch of the conditional type. It is possible to have multiple infer locations for the same type variable.

For example, the following extracts the return type of a function type:

ts
type ReturnType<T> = T extends (...args: any[]) => infer R ? R : any;

Conditional types can be nested to form a sequence of pattern matches that are evaluated in order:

ts
type Unpacked<T> = T extends (infer U)[] ? U : T extends (...args: any[]) => infer U ? U : T extends Promise<infer U> ? U : T; type T0 = Unpacked<string>; // string type T1 = Unpacked<string[]>; // string type T2 = Unpacked<() => string>; // string type T3 = Unpacked<Promise<string>>; // string type T4 = Unpacked<Promise<string>[]>; // Promise<string> type T5 = Unpacked<Unpacked<Promise<string>[]>>; // string

The following example demonstrates how multiple candidates for the same type variable in co-variant positions causes a union type to be inferred:

ts
type Foo<T> = T extends { a: infer U; b: infer U } ? U : never; type T10 = Foo<{ a: string; b: string }>; // string type T11 = Foo<{ a: string; b: number }>; // string | number

Likewise, multiple candidates for the same type variable in contra-variant positions causes an intersection type to be inferred:

ts
type Bar<T> = T extends { a: (x: infer U) => void; b: (x: infer U) => void } ? U : never; type T20 = Bar<{ a: (x: string) => void; b: (x: string) => void }>; // string type T21 = Bar<{ a: (x: string) => void; b: (x: number) => void }>; // string & number

When inferring from a type with multiple call signatures (such as the type of an overloaded function), inferences are made from the last signature (which, presumably, is the most permissive catch-all case). It is not possible to perform overload resolution based on a list of argument types.

ts
declare function foo(x: string): number; declare function foo(x: number): string; declare function foo(x: string | number): string | number; type T30 = ReturnType<typeof foo>; // string | number

It is not possible to use infer declarations in constraint clauses for regular type parameters:

ts
type ReturnType<T extends (...args: any[]) => infer R> = R; // Error, not supported

However, much the same effect can be obtained by erasing the type variables in the constraint and instead specifying a conditional type:

ts
type AnyFunction = (...args: any[]) => any; type ReturnType<T extends AnyFunction> = T extends (...args: any[]) => infer R ? R : any;

Predefined conditional types

TypeScript 2.8 adds several predefined conditional types to lib.d.ts:

  • Exclude<T, U> — Exclude from T those types that are assignable to U.
  • Extract<T, U> — Extract from T those types that are assignable to U.
  • NonNullable<T> — Exclude null and undefined from T.
  • ReturnType<T> — Obtain the return type of a function type.
  • InstanceType<T> — Obtain the instance type of a constructor function type.

Example

ts
type T00 = Exclude<"a" | "b" | "c" | "d", "a" | "c" | "f">; // "b" | "d" type T01 = Extract<"a" | "b" | "c" | "d", "a" | "c" | "f">; // "a" | "c" type T02 = Exclude<string | number | (() => void), Function>; // string | number type T03 = Extract<string | number | (() => void), Function>; // () => void type T04 = NonNullable<string | number | undefined>; // string | number type T05 = NonNullable<(() => string) | string[] | null | undefined>; // (() => string) | string[] function f1(s: string) { return { a: 1, b: s }; } class C { x = 0; y = 0; } type T10 = ReturnType<() => string>; // string type T11 = ReturnType<(s: string) => void>; // void type T12 = ReturnType<<T>() => T>; // {} type T13 = ReturnType<<T extends U, U extends number[]>() => T>; // number[] type T14 = ReturnType<typeof f1>; // { a: number, b: string } type T15 = ReturnType<any>; // any type T16 = ReturnType<never>; // never type T17 = ReturnType<string>; // Error type T18 = ReturnType<Function>; // Error type T20 = InstanceType<typeof C>; // C type T21 = InstanceType<any>; // any type T22 = InstanceType<never>; // never type T23 = InstanceType<string>; // Error type T24 = InstanceType<Function>; // Error

Note: The Exclude type is a proper implementation of the Diff type suggested here. We’ve used the name Exclude to avoid breaking existing code that defines a Diff, plus we feel that name better conveys the semantics of the type.

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Last updated: Jul 04, 2020