# A Look at TypeScript's Conditional Types

published on March 11th, 2018

TypeScript 2.8 introduces a fantastic new feature called conditional types which is going to make TypeScript's type system even more powerful and enable a huge range of type orperators which were previously not possible. Let's take a look at how this exciting new feature works and what we can do with it.

Conditional types essentially give you an if statement along with the ability to ask questions at the type level. This enables some very powerful constructs which were not possible before. Conditional types look as follows:

    T extends U ? X : Y


In Anders Hejlsberg's words this adds the ability to express non-uniform type mappings. Before conditional types, there was basically no way to make an either-or decision in type operators, now there is.

The question in the conditional type is the extends check - if T extends U then the resulting type is X, otherwise it is Y. A type extends another type if it is assignable to it, or in some sense "at least as big as" the other type. If T extends U you can do everything with T that you can do with U or use T everywhere you can use U.

Here is a very simple (and slightly silly) example for using conditional types:

type IsNumberType<T> = T extends number ? "yes" : "no";

type Yes = IsNumberType<3>; // type "yes"
type No = IsNumberType<"foo">; // type "no"


A more interesting and more useful example is to implement JavaScript's typeof operator at TypeScript's type level:

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"


## The Distributive Rule of Conditional and Union Types

One interesting rule about conditional types is how they interact with union types. A conditional types distributes over a union type with the following distribution law:

(A | B) extends T ? X : U = (A extends T ? X : U) | (B extends T ? X : U)


Let's see how we can apply this law to the TypeName operator defined above:

TypeName<string | (() => void)>
= (string | (() => void)) extends string ? "string" :
(string | (() => void)) extends number ? "number" :
(string | (() => void)) extends boolean ? "boolean" :
(string | (() => void)) extends undefined ? "undefined" :
(string | (() => void)) extends Function ? "function" :
"object";
= (string extends string ? "string" :
string extends number ? "number" :
string extends boolean ? "boolean" :
string extends undefined ? "undefined" :
string extends Function ? "function" :
"object")
|
((() => void) extends string ? "string" :
(() => void) extends number ? "number" :
(() => void) extends boolean ? "boolean" :
(() => void) extends undefined ? "undefined" :
(() => void) extends Function ? "function" :
"object")
= "string" | "function"


So we have a substitution rule by which we can evaluate conditional types where the argument is a union type.

### The Diff Type Operator

An interesting and useful application of this distribution rule is to remove cases from union types. For example, it is easy to write a Diff type operator which relies on this rule:

type Diff<T, U> = T extends U ? never : T;


This might seem confusing at first (it certainly did to me!), but makes sense when looking at an example:

Diff<"a" | "b" | "c", "a" | "b">
= ("a" | "b" | "c") extends ("a" | "b") ? never : ("a" | "b" | "c")
=   "a" extends ("a" | "b") ? never : "a"
| "b" extends ("a" | "b") ? never : "b"
| "c" extends ("a" | "b") ? never : "c"
=   never
| never
| "c"
= "c"


### The NonNullable Type Operator

A useful application to this the NonNullable type operator which removes the types null and undefined from a union type:

type NonNullable<T> = Diff<T, null | undefined>;

type Foo = NonNullable<string | null | undefined>; // Foo = string;


It would seem like we could the inverse of the Partial type operator with this technique - namely making all parameters on an object required. However, this is not the case:

type AlmostNonPartial<T extends object> =
{ [KEY in keyof T]: NonNullable<T[KEY]>; };

type Bar = AlmostNonPartial<{ a?: number; b?: string; }>;
// o_O - a and b are still optional on Bar:
// Bar = { a?: number | undefined; b?: string | undefined; }


Fear not - rescue is in sight, TypeScript 2.8 also provides [better control over mapped type modifiers], with which we can write the NonPartial operator:

// notice the -? which removes the ? modifier
type NonPartial<T extends object> =
{ [KEY in keyof T]-?: T[KEY]; };

type Bar = NonPartial<{ a?: number; b?: string; }>;
// Bar now has the right type:
// Bar = { a: number; b: string; }


## Leveraging Recursive Types to implement DeepReadonly

In TypeScript type mappings can be recursive under certain conditions. Using recursive type definitions and conditional types, we can implement the coveted DeepReadonly type operator with conditional and recursive types:

type DeepReadonly<T> =
T extends any[] ? DeepReadonlyArray<T[number]> :
T extends object ? DeepReadonlyObject<T> :
T;