A string.

The function's name property can be used to identify the function in debugging tools or error messages. It has no semantic significance to the language itself.

The name property is read-only and cannot be changed by the assignment operator:

function someFunction() {}

someFunction.name = "otherFunction";
console.log(someFunction.name); // someFunction

To change it, use Object.defineProperty().

The name property is typically inferred from how the function is defined. In the following sections, we will describe the various ways in which it can be inferred.

The name property returns the name of a function declaration.

function doSomething() {}
doSomething.name; // "doSomething"

An export default declaration exports the function as a declaration instead of an expression. If the declaration is anonymous, the name is "default".

// -- someModule.js --
export default function () {}

// -- main.js --
import someModule from "./someModule.js";

someModule.name; // "default"

Functions created with the Function() constructor have name "anonymous".

new Function().name; // "anonymous"

If the function expression is named, that name is used as the name property.

const someFunction = function someFunctionName() {};
someFunction.name; // "someFunctionName"

Anonymous function expressions created using the keyword function or arrow functions would have "" (an empty string) as their name.

(function () {}).name; // ""
(() => {}).name; // ""

However, such cases are rare — usually, in order to refer to the expression elsewhere, the function expression is attached to an identifier when it's created (such as in a variable declaration). In such cases, the name can be inferred, as the following few subsections demonstrate.

One practical case where the name cannot be inferred is a function returned from another function:

function getFoo() {
  return () => {};
}
getFoo().name; // ""

Variables and methods can infer the name of an anonymous function from its syntactic position.

const f = function () {};
const object = {
  someMethod: function () {},
};

console.log(f.name); // "f"
console.log(object.someMethod.name); // "someMethod"

The same applies to assignment:

let f;
f = () => {};
f.name; // "f"

Functions in initializers (default values) of destructuring, default parameters, class fields, etc., will inherit the name of the bound identifier as their name.

const [f = () => {}] = [];
f.name; // "f"

const { someMethod: m = () => {} } = {};
m.name; // "m"

function foo(f = () => {}) {
  console.log(f.name);
}
foo(); // "f"

class Foo {
  static someMethod = () => {};
}
Foo.someMethod.name; // someMethod

Function.prototype.bind() produces a function whose name is "bound " plus the function name.

function foo() {}
foo.bind({}).name; // "bound foo"

When using get and set accessor properties, "get" or "set" will appear in the function name.

const o = {
  get foo() {},
  set foo(x) {},
};

const descriptor = Object.getOwnPropertyDescriptor(o, "foo");
descriptor.get.name; // "get foo"
descriptor.set.name; // "set foo";

A class's name follows the same algorithm as function declarations and expressions.

class Foo {}
Foo.name; // "Foo"

If a Symbol is used a function name and the symbol has a description, the method's name is the description in square brackets.

const sym1 = Symbol("foo");
const sym2 = Symbol();

const o = {
  [sym1]() {},
  [sym2]() {},
};

o[sym1].name; // "[foo]"
o[sym2].name; // "[]"

Private fields and private methods have the hash (#) as part of their names.

class Foo {
  #field = () => {};
  #method() {}
  getNames() {
    console.log(this.#field.name);
    console.log(this.#method.name);
  }
}

new Foo().getNames();
// "#field"
// "#method"

You can use obj.constructor.name to check the "class" of an object.

function Foo() {} // Or: class Foo {}

const fooInstance = new Foo();
console.log(fooInstance.constructor.name); // "Foo"

However, because static members will become own properties of the class, we can't obtain the class name for virtually any class with a static method property name():

class Foo {
  constructor() {}
  static name() {}
}

With a static name() method Foo.name no longer holds the actual class name but a reference to the name() function object. Trying to obtain the class of fooInstance via fooInstance.constructor.name won't give us the class name at all, but instead a reference to the static class method. Example:

const fooInstance = new Foo();
console.log(fooInstance.constructor.name); // ƒ name() {}

Due to the existence of static fields, name may not be a function either.

class Foo {
  static name = 123;
}
console.log(new Foo().constructor.name); // 123

If a class has a static property called name, it will also become writable. The built-in definition in the absence of a custom static definition is read-only:

Foo.name = "Hello";
console.log(Foo.name); // "Hello" if class Foo has a static "name" property, but "Foo" if not.

Therefore you may not rely on the built-in name property to always hold a class's name.

Source code such as:

function Foo() {}
const foo = new Foo();

if (foo.constructor.name === "Foo") {
  console.log("'foo' is an instance of 'Foo'");
} else {
  console.log("Oops!");
}

may be compressed to:

function a() {}
const b = new a();
if (b.constructor.name === "Foo") {
  console.log("'foo' is an instance of 'Foo'");
} else {
  console.log("Oops!");
}

In the uncompressed version, the program runs into the truthy branch and logs "'foo' is an instance of 'Foo'" — whereas, in the compressed version it behaves differently, and runs into the else branch. If you rely on the name property, like in the example above, make sure your build pipeline doesn't change function names, or don't assume a function has a particular name.

• Polyfill for Function: name in core-js
• Function