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How do you declare a function with a generic parameter?

TypeScriptJunior~1 min read
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Basic syntax

javascript
function name<T>(arg: T): T { return arg; }
  • T is the generic parameter.
  • When the function is called, TypeScript substitutes the required type for T on its own.

Example 1: a simple generic function

javascript
function identity<T>(value: T): T { return value; } const num = identity(42); // T = number const str = identity("hello"); // T = string

TypeScript understands that num is number, and str is string.

Example 2: explicit type argument when calling

javascript
function wrapValue<T>(value: T): { data: T } { return { data: value }; } const wrapped = wrapValue<string>("text"); // { data: string }

Here we manually specified the type <string>, although TypeScript usually does this automatically.

Example 3: multiple generic parameters

javascript
function merge<T, U>(a: T, b: U): T & U { return Object.assign({}, a, b); } const person = merge({ name: "Tim" }, { age: 25 }); // person has the type { name: string; age: number }

You can use several generic parameters (<T, U, V, ...>).

Example 4: with a constraint (extends)

javascript
function getLength<T extends { length: number }>(value: T): number { return value.length; } getLength("hello"); // a string has length getLength([1, 2, 3]); // an array has length getLength(123); // a number does not have length

extends constrains which types can be substituted for T.

Example 5: with arrays

javascript
function firstElement<T>(arr: T[]): T { return arr[0]; } const first = firstElement([1, 2, 3]); // number const second = firstElement(["a", "b"]); // string

TypeScript infers the type of the array elements automatically.

Summary

SyntaxExampleDescription
<T>function foo<T>(x: T): TOne generic parameter
<T, U>function bar<T, U>(a: T, b: U)Several generic parameters
<T extends SomeType>function baz<T extends object>(x: T)A type constraint
<T = DefaultType>function qux<T = string>(x?: T)A default value for the generic

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