Why are derived class property values not seen in the base class constructor?
I wrote some code:
class Base {
// Default value
myColor = 'blue';
constructor() {
console.log(this.myColor);
}
}
class Derived extends Base {
myColor = 'red';
}
// Prints "blue", expected "red"
const x = new Derived();
I was expecting my derived class field initializer to run before the base class constructor.
Instead, the derived class doesn't change the myColor property until after the base class constructor runs, so I observe the wrong values in the constructor.
Is this a bug? What's wrong? Why does this happen? What should I do instead?
Not a Bug
First up, this is not a bug in TypeScript, Babel, or your JS runtime.
Why It Has To Be This Way
The first follow-up you might have is "Why not do this correctly!?!?". Let's examine the specific case of TypeScript emit. The actual answer depends on what version of ECMAScript we're emitting class code for.
Downlevel emit: ES3/ES5
Let's examine the code emitted by TypeScript for ES3 or ES5. I've simplified + annotated this a bit for readability:
var Base = (function () {
function Base() {
// BASE CLASS PROPERTY INITIALIZERS
this.myColor = 'blue';
console.log(this.myColor);
}
return Base;
}());
var Derived = (function (_super) {
__extends(Derived, _super);
function Derived() {
// RUN THE BASE CLASS CTOR
_super();
// DERIVED CLASS PROPERTY INITIALIZERS
this.myColor = 'red';
// Code in the derived class ctor body would appear here
}
return Derived;
}(Base));
The base class emit is uncontroversially correct - the fields are initialized, then the constructor body runs. You certainly wouldn't want the opposite - initializing the fields before running the constructor body would mean you couldn't see the field values until after the constructor, which is not what anyone wants.
Is the derived class emit correct?
No, you should swap the order
Many people would argue that the derived class emit should look like this:
// DERIVED CLASS PROPERTY INITIALIZERS
this.myColor = 'red';
// RUN THE BASE CLASS CTOR
_super();
This is super wrong for any number of reasons:
- It has no corresponding behavior in ES6 (see next section)
- The value
'red'formyColorwill be immediately overwritten by the base class value 'blue' - The derived class field initializer might invoke base class methods which depend on base class initializations.
On that last point, consider this code:
class Base {
thing = 'ok';
getThing() { return this.thing; }
}
class Derived extends Base {
something = this.getThing();
}
If the derived class initializers ran before the base class initializers, Derived#something would always be undefined, when clearly it should be 'ok'.
No, you should use a time machine
Many other people would argue that a nebulous something else should be done so that Base knows that Derived has a field initializer.
You can write example solutions that depend on knowing the entire universe of code to be run. But TypeScript / Babel / etc cannot guarantee that this exists. For example, Base can be in a separate file where we can't see its implementation.
Downlevel emit: ES6
If you didn't already know this, it's time to learn: classes are not a TypeScript feature. They're part of ES6 and have defined semantics. But ES6 classes don't support field initializers, so they get transformed to ES6-compatible code. It looks like this:
class Base {
constructor() {
// Default value
this.myColor = 'blue';
console.log(this.myColor);
}
}
class Derived extends Base {
constructor() {
super(...arguments);
this.myColor = 'red';
}
}
Instead of
super(...arguments);
this.myColor = 'red';
Should we have this?
this.myColor = 'red';
super(...arguments);
No, because it doesn't work. It's illegal to refer to this before invoking super in a derived class. It simply cannot work this way.
ES7+: Public Fields
The TC39 committee that controls JavaScript is investigating adding field initializers to a future version of the language.
You can read about it on GitHub or read the specific issue about initialization order.
OOP refresher: Virtual Behavior from Constructors
All OOP languages have a general guideline, some enforced explicitly, some implicitly by convention:
Do not call virtual methods from the constructor
Examples:
- C# Virtual member call in a constructor
- C++ Calling virtual functions inside constructors
- Python Calling member functions from a constructor
- Java Is it OK to call abstract method from constructor in Java?
In JavaScript, we have to expand this rule a little
Do not observe virtual behavior from the constructor
and
Class property initialization counts as virtual
Solutions
The standard solution is to transform the field initialization to a constructor parameter:
class Base {
myColor: string;
constructor(color: string = "blue") {
this.myColor = color;
console.log(this.myColor);
}
}
class Derived extends Base {
constructor() {
super("red");
}
}
// Prints "red" as expected
const x = new Derived();
You can also use an init pattern, though you need to be cautious to not observe virtual behavior from it and to not do things in the derived init method that require a complete initialization of the base class:
class Base {
myColor: string;
constructor() {
this.init();
console.log(this.myColor);
}
init() {
this.myColor = "blue";
}
}
class Derived extends Base {
init() {
super.init();
this.myColor = "red";
}
}
// Prints "red" as expected
const x = new Derived();
I would respectfully argue this is, in fact, a bug
By doing an unexpected thing, this is undesired behavior that breaks common class extension use cases. Here is the initialization order that would support your use case and that I would argue is better:
Base property initializers
Derived property initializers
Base constructor
Derived constructor
Problems / Solutions
- The typescript compiler currently emits property initializations in the constructor
The solution here is to separate the property initializations from the calling of the constructor functions. C# does this, although it inits base properties after derived properties, which is also counterintuitive. This could be accomplished by emitting helper classes so that the derived class can initialize the base class in an arbitrary order.
class _Base {
ctor() {
console.log('base ctor color: ', this.myColor);
}
initProps() {
this.myColor = 'blue';
}
}
class _Derived extends _Base {
constructor() {
super();
}
ctor() {
super.ctor();
console.log('derived ctor color: ', this.myColor);
}
initProps() {
super.initProps();
this.myColor = 'red';
}
}
class Base {
constructor() {
const _class = new _Base();
_class.initProps();
_class.ctor();
return _class;
}
}
class Derived {
constructor() {
const _class = new _Derived();
_class.initProps();
_class.ctor();
return _class;
}
}
// Prints:
// "base ctor color: red"
// "derived ctor color: red"
const d = new Derived();
- Won't the base constructor break because we're using derived class properties?
Any logic that breaks in the base constructor can be moved to a method that would be overridden in the derived class. Since derived methods are initialized before the base constructor is called, this would work correctly. Example:
class Base {
protected numThings = 5;
constructor() {
console.log('math result: ', this.doMath())
}
protected doMath() {
return 10/this.numThings;
}
}
class Derived extends Base {
// Overrides. Would cause divide by 0 in base if we weren't overriding doMath
protected numThings = 0;
protected doMath() {
return 100 + this.numThings;
}
}
// Should print "math result: 100"
const x = new Derived();