问题
Given a simple class
class Foo {
constructor(x) {
if (!(this instanceof Foo)) return new Foo(x);
this.x = x;
}
hello() {
return `hello ${this.x}`;
}
}
Is it possible to call the class constructor without the new keyword?
Usage should allow
(new Foo(\"world\")).hello(); // \"hello world\"
Or
Foo(\"world\").hello(); // \"hello world\"
But the latter fails with
Cannot call a class as a function
回答1:
Classes have a "class body" that is a constructor.
If you use an internal constructor() function, that function would be the same class body as well, and would be what is called when the class is called, hence a class is always a constructor.
Constructors requires the use of the new operator to create a new instance, as such invoking a class without the new operator results in an error, as it's required for the class constructor to create a new instance.
The error message is also quite specific, and correct
TypeError: Class constructors cannot be invoked without 'new'
You could;
- either use a regular function instead of a class1.
- Always call the class with
new. - Call the class inside a wrapping regular function, always using
new, that way you get the benefits of classes, but the wrapping function can still be called with and without thenewoperator2.
1)
function Foo(x) {
if (!(this instanceof Foo)) return new Foo(x);
this.x = x;
this.hello = function() {
return this.x;
}
}
2)
class Foo {
constructor(x) {
this.x = x;
}
hello() {
return `hello ${this.x}`;
}
}
var _old = Foo;
Foo = function(...args) { return new _old(...args) };
回答2:
As others have pointed out ES2015 spec strictly states that such call should throw TypeError, but at the same time it provides feature that can be used to achieve exactly the desired result, namely Proxies.
Proxies allows us to virtualize over a concept of an object. For instance they can be used to change some behaviour of particular object without affecting anything else.
In your specific use case class Foo is Function object which can be called -- this normally means that body of this function will be executed. But this can be changed with Proxy:
const _Foo = new Proxy(Foo, {
// target = Foo
apply (target, thisArg, argumentsList) {
return new target(...argumentsList);
}
});
_Foo("world").hello();
const f = _Foo("world");
f instanceof Foo; // true
f instanceof _Foo; // true
(Note that _Foo is now the class you want to expose, so identifiers should probably be the other way round)
If run by browser that support Proxies, calling _Foo(...) will now execute apply trap function instead of the orignal constructor.
At the same time this "new" _Foo class is indistinguishable from original Foo (apart from being able to call it as a normal function). Similarily there is no difference by which you can tell object created with Foo and _Foo.
The biggest downside of this is that it cannot be transpilled or pollyfilled, but still its viable solution for having Scala-like class apply in JS in the future.
回答3:
Here's a pattern I've come across that really helps me. It doesn't use a class, but it doesn't require the use of new either. Win/Win.
const Foo = x => ({
x,
hello: () => `hello ${x}`,
increment: () => Foo(x + 1),
add: ({x: y}) => Foo(x + y)
})
console.log(Foo(1).x) // 1
console.log(Foo(1).hello()) // hello 1
console.log(Foo(1).increment().hello()) // hello 2
console.log(Foo(1).add(Foo(2)).hello()) // hello 3回答4:
No, this is not possible. Constructors that are created using the class keyword can only be constructed with new, if they are [[call]]ed without they always throw a TypeError1 (and there's not even a way to detect this from the outside).
1: I'm not sure whether transpilers get this right
You can use a normal function as a workaround, though:
class Foo {
constructor(x) {
this.x = x;
}
hello() {
return `hello ${this.x}`;
}
}
{
const _Foo = Foo;
Foo = function(...args) {
return new _Foo(...args);
};
Foo.prototype = _Foo.prototype;
}
Disclaimer: instanceof and extending Foo.prototype work as normal, Foo.length does not, .constructor and static methods do not but can be fixed by adding Foo.prototype.constructor = Foo; and Object.setPrototypeOf(Foo, _Foo) if required.
For subclassing Foo (not _Foo) with class Bar extends Foo …, you should use return Reflect.construct(_Foo, args, new.target) instead of the new _Foo call. Subclassing in ES5 style (with Foo.call(this, …)) is not possible.
回答5:
class MyClass {
constructor(param) {
// ...
}
static create(param) {
return new MyClass(param);
}
doSomething() {
// ...
}
}
MyClass.create('Hello World').doSomething();
Is that what you want?
If you need some logic when creating a new instance of MyClass, it could be helpful to implement a "CreationStrategy", to outsorce the logic:
class MyClassCreationStrategy {
static create(param) {
let instance = new MyClass();
if (!param) {
// eg. handle empty param
}
instance.setParam(param);
return instance;
}
}
class DefaultCreationStrategy {
static create(classConstruct) {
return new classConstruct();
}
}
MyClassCreationStrategy.create(param).doSomething();
DefaultCreationStrategy.create(MyClass).doSomething();
回答6:
i just made this npm module for you ;)
https://www.npmjs.com/package/classy-decorator
import classy from "classy-decorator";
@classy()
class IamClassy {
constructor() {
console.log("IamClassy Instance!");
}
}
console.log(new IamClassy() instanceof IamClassy); // true
console.log(IamClassy() instanceof IamClassy); // true
回答7:
Dug up this one in the draft
Constructors defined using class definition syntax throw when called as functions
So I guess that's not possible with classes.
回答8:
Call class constructor manually can be usefull when refactoring code (having parts of the code in ES6, other parts beeing function & prototype definition)
I ended up with a small, yet usefull boilerplate, slicing the constructor into another function. Period.
class Foo {
constructor() {
//as i will not be able to call the constructor, just move everything to initialize
this.initialize.apply(this, arguments)
}
initialize() {
this.stuff = {};
//whatever you want
}
}
function Bar () {
Foo.prototype.initialize.call(this);
}
Bar.prototype.stuff = function() {}
回答9:
Here's a where you can use a 'scope safe constructor' Observe this code:
function Student(name) {
if(this instanceof Student) {
this.name = name;
} else {
return new Student(name);
}
}
Now you can create a Student object without using new as follows:
var stud1 = Student('Kia');
回答10:
I had problems extending classes converted with the transformation function mentioned in some other answers. The issue seems to be that node (as of v9.4.0) doesn't properly support the argument spread operator ((...args) =>).
This function based on the transpiled output of the classy-decorator (mentioned in another answer) works for me and doesn't require support for decorators or the argument spread operator.
// function that calls `new` for you on class constructors, simply call
// YourClass = bindNew(YourClass)
function bindNew(Class) {
function _Class() {
for (
var len = arguments.length, rest = Array(len), key = 0;
key < len;
key++
) {
rest[key] = arguments[key];
}
return new (Function.prototype.bind.apply(Class, [null].concat(rest)))();
}
_Class.prototype = Class.prototype;
return _Class;
}
Usage:
class X {}
X = bindNew(X);
// or
const Y = bindNew(class Y {});
const x = new X();
const x2 = X(); // woohoo
x instanceof X; // true
x2 instanceof X; // true
class Z extends X {} // works too
As a bonus, TypeScript (with "es5" output) seems to be fine with the old instanceof trick (well, it won't typecheck if used without new but it works anyhow):
class X {
constructor() {
if (!(this instanceof X)) {
return new X();
}
}
}
because it compiles it down to:
var X = /** @class */ (function () {
function X() {
if (!(this instanceof X)) {
return new X();
}
}
return X;
}());
回答11:
Alright I have another answer here, and I think this one is pretty innovative.
Basically, the problem with doing something similar to Naomik's answer is that you create functions each and every time you chain methods together.
EDIT: This solution shares the same problem, however, this answer is being left up for educational purposes.
So here I'm offering a way to merely bind new values to your methods--which are basically just independent functions. This offer the additional benefit of being able to import functions from different modules into the newly constructed object.
Okay, so here it goes.
const assoc = (prop, value, obj) =>
Object.assign({},obj,{[prop]: value})
const reducer = ( $values, accumulate, [key,val] ) => assoc( key, val.bind( undefined,...$values ), accumulate )
const bindValuesToMethods = ( $methods, ...$values ) =>
Object.entries( $methods ).reduce( reducer.bind( undefined, ...$values), {} )
const prepareInstance = (instanceMethods, staticMethods = ({}) ) => Object.assign(
bindValuesToMethods.bind( undefined, instanceMethods ),
staticMethods
)
// Let's make our class-like function
const RightInstanceMethods = ({
chain: (x,f) => f(x),
map: (x,f) => Right(f(x)),
fold: (x,l,r) => r(x),
inspect: (x) => `Right(${x})`
})
const RightStaticMethods = ({
of: x => Right(x)
})
const Right = prepareInstance(RightInstanceMethods,RightStaticMethods)
Now you can do
Right(4)
.map(x=>x+1)
.map(x=>x*2)
.inspect()
You can also do
Right.of(4)
.map(x=>x+1)
.map(x=>x*2)
.inspect()
You also have the added benefit of being able to export from modules as such
export const Right = prepareInstance(RightInstanceMethods,RightStaticMethods)
While you don't get ClassInstance.constructor you do have FunctorInstance.name (note, you may need to polyfill Function.name and/or not use an arrow function for export for browser compatibility with Function.name purposes)
export function Right(...args){
return prepareInstance(RightInstanceMethods,RightStaticMethods)(...args)
}
PS - New name suggestions for prepareInstance welcomed, see Gist.
https://gist.github.com/babakness/56da19ba85e0eaa43ae5577bc0064456
回答12:
Calling the class constructor without the new keyword is not possible.
The error message is quite specific.
See a blog post on 2ality and the spec:
However, you can only invoke a class via new, not via a function call (Sect. 9.2.2 in the spec):
> Point()
TypeError: Classes can’t be function-called
回答13:
I'm adding this as a follow up to a comment by naomik and utilizing on the method illustrated by Tim and Bergi. I'm also going to suggest an of function to use as a general case.
To do this in a functional way AND utilize the efficiency of prototypes (not re-create all method each time a new instance is created), one could use this pattern
const Foo = function(x){ this._value = x ... }
Foo.of = function(x){ return new Foo(x) }
Foo.prototype = {
increment(){ return Foo.of(this._value + 1) },
...
}
Please note that this is consistent with fantasy-land JS specs
https://github.com/fantasyland/fantasy-land#of-method
I personally feel that it is cleaner to use the ES6 class syntax
class Foo {
static of(x) { new Foo(x)}
constructor(x) { this._value = x }
increment() { Foo.of(this._value+1) }
}
Now one could wrap this in a closure as such
class Foo {
static of(x) { new _Foo(x)}
constructor(x) { this._value = x }
increment() { Foo.of(this._value+1) }
}
function FooOf (x) {
return Foo.of(x)
}
Or rename FooOf and Foo as desired, ie the class could be FooClass and the function just Foo, etc.
This is better than place the class in the function because creating new instances doesn't burden us with creating new classes as well.
Yet another way is to create a an of function
const of = (classObj,...args) => (
classObj.of
? classObj.of(value)
: new classObj(args)
)
And then do something like of(Foo,5).increment()
回答14:
This might be a little contrived, but it works
function Foo(x){
"use strict"
class Bar {
constructor(x) {
if (!(this instanceof Bar)) return new Bar(x);
this.x = x;
}
hello() {
return `hello ${this.x}`;
}
}
return new Bar(x)
}
Foo("world").hello()
回答15:
You can't use a class without the new constructor, in my case I didn't want to use the new constructor any time I wanted to use my class, so what you can do is to wrap your class as follows (in my case it's a Dates utils library):
来源：https://stackoverflow.com/questions/30689817/es6-call-class-constructor-without-new-keyword