Questioning the mysterious reference error in the JS class from MDN page. The structure of the Bad class's constructor leaves me baffled – is it because the empty constructor calls super() as a default?
class Base {}
class Good extends Base {}
class AlsoGood extends Base {
constructor() {
return {a: 5};
}
}
class Bad extends Base {
constructor() {}
}
new Good();
new AlsoGood();
new Bad(); // ReferenceError
In order to prevent encountering this issue, you have two options: either eliminate the constructor from the class entirely, or ensure that super() is called within the Bad class constructor.
Is it because the default behavior of the empty constructor in the Bad class is to call `super()`?
Actually, no - if the empty constructor automatically called `super`, you wouldn't encounter the error.
For a derived class's constructor to be valid, it must manually invoke the parent's constructor using `super`.
In cases where a derived class lacks any explicit constructor, `super` will still be invoked automatically, like so:
class Good extends Base {
}
This effectively translates to the following equivalent code block:
class Good extends Base {
constructor(...args) {
super(...args);
}
}
The rule about automatic inheritance through empty constructors does not apply. If the constructor is defined, `super` will not be automatically called - however, it must be included in the constructor logic.
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