In the TypeScript handbook, when it mentions "can be considered as the interface type," what exactly is meant by that?

The manual emphasizes that:

It’s crucial to understand that an implements clause is merely a confirmation that the class can be used as if it were of the interface type. It does not alter the class's type or methods in any way. One common mistake is assuming that by using an implements clause, the class type will change - this is not the case!

I'm confused about what is meant by "can be treated as the interface type" when it also states, "It doesn’t change the type of the class or its methods at all."

Does this imply that I have to specify identical types for both the class and the interface?

An example they provide involves the variable s being of type any:

interface Checkable {
  check(name: string): boolean;
}
 
class NameChecker implements Checkable {
  check(s) {
    // No error produced here
    return s.toLowerCase() === "ok";
  }
}

Answer №1

Section 1: Treatable as a type of interface:

Imagine you define an interface and have two different classes implement it as shown below:

interface Alphabet {
  x: number;
  y?: number;
}
class C implements Alphabet { //Error
  z = 0; 
}
class D implements Alphabet { 
  x= 1; 
  y = 0;
}

c triggers an error because it cannot be regarded as something of the type Alphabet. The use of the implements keyword is to enforce this check and throw an error if not adhered to.

Playground

Section 2: No change in the class or its methods' types whatsoever:

In the following example, one might mistakenly assume that c.y would be valid, even though undefined. However, that is not true. Since the type of c remains C, it does not become Alphabet.

interface Alphabet {
  x: number;
  y?: number;
}
class C implements Alphabet {
  x = 0;
}
class D implements Alphabet {
  x= 1;
  y = 0;
}
const c = new C();
const d = new D();

c.y; //Error
d.y; //No Error

Playground

And this is where the confusion lies, as stated in the handbook.

Answer №2

In Typescript, the implements keyword plays a key role in ensuring structural subtyping is maintained.

When a class implements an interface, instances of that class can be treated as instances of the interface, allowing for seamless structural subtyping.

However, there may be confusion about how a class "can be treated as the interface type" while still retaining its original type and methods as stated: "It doesn’t change the type of the class or its methods at all."

Unlike traditional inheritance where subclasses become subtypes of the parent class and inherit its methods, implementing an interface simply guarantees that an instance of the class can be used as an instance of the interface if it fully adheres to the specified contract through complete implementation.

This distinction becomes more apparent when we look at code examples and see how TypeScript handles method signatures within implemented interfaces.


The provided example with the code snippet check(s) { ... } might appear incomplete in terms of types but satisfies the necessary contract requirements nonetheless.

To elaborate further on this scenario:

Let's consider the following interface:

interface Checkable {
  check(<b>S: Type</b>): boolean;
}

Two classes, NameChecker and AgeChecker, implement this interface with distinct method signatures:

class NameChecker implements Checkable {
  check(<b>name: string</b>): boolean
  {
    return name.toLowerCase() === "john doe";
  }
}

class AgeChecker implements Checkable {
  check(<b>age: number</b>): boolean
  {
    return age === 50;
  }
}

Despite the differences in method signatures, the above implementations are deemed valid by the implements keyword due to the principles of structural typing upheld by TypeScript. This allows for flexibility in function declarations as long as they satisfy the contract outlined in the corresponding interface.


[1] While Typescript already leverages structural subtyping, the inclusion of the implements keyword enforces strict adherence to the specified contract, hence the emphasis on "ensures" in the explanation.

[2] It is essential to note that the omission of specific types leads to implicit typing as any.

Answer №3

"Being recognized as the interface type" means that whenever the implemented interface Checkable is needed or expected in the codebase, an instance of the implementing NameChecker will perfectly fulfill that requirement.

"It doesn't alter the class's type" in the sense that through implementing the interface, the class and its type remain unchanged and will not be adapted to match the interface's methods more closely. Either they align or they do not, resulting in an error being thrown and preventing the code from compiling.

Implementing the interface necessitates all properties and methods of the interface to also be present in the class definition. Thus, you will indeed need to specify the same types for both. The interface exclusively outlines the types (methods, arguments, and return type), while the class includes the specific implementation of each method.

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