What's the best way to address this blind spot?

Exploring the world of TypeScript has left me puzzled by a scenario where TypeScript does not perform type checking as expected. I'm running into an issue where 'this.a.method()' appears to be error-free when it should actually throw an error. Ideally, TypeScript would alert me that a.method() needs a parameter value of type string. What might be the factor causing this discrepancy?

class A {
  method(value: string){
    console.log(value);
  }
}
class B {
  constructor(private a){}
  call(){
    this.a.method()
  }
}
const a = new A();
const b = new B(a);
b.call();

Answer №1

In order to specify the type for private a; within B's constructor, you would need to write it as constructor(private a: A){}

class A {
  method(value: string){
    console.log(value);
  }
}
class B {
  constructor(private a: A){}
  call(){
    this.a.method()
  }
}
const a = new A();
const b = new B(a);
b.call();

If you fail to provide an argument when calling this.a.method(), you will encounter a compilation error which can be resolved by passing a string as an argument like so: this.a.method("bar").

Answer №2

When you fail to specify a type for a variable or property, it defaults to any, which essentially signifies no need for type validation. This differs from type object. To enforce type checking in TypeScript, assign a specific type to your property like a with

constructor(private a: SomeType) { }

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