method-signature-style with overloads, what's the recommended practice?

[yf-yang]

Related: #2706, #4106

I enabled method-signature-style, here is an example

interface testInterface {
  // first style
  // foo(a: number): number;
  // foo(a: string): string;
  // foo(a: number | string): number | string;

  // second style
  // foo: ((a: number) => number) & ((a: string) => string) & ((a: number|string) => number|string);

  // third style
  // foo: {
  //     (a: number): number;
  //     (a: string): string;
  //     (a: number | string): number | string;
  // }
}

class TestClass implements testInterface {
  foo(a: number): number;
  foo(a: string): string;
  foo(a: number | string) {
    return a;
  }
}

function test(a: number | string, c: testInterface) {
  const b = c.foo(a);
}

The first style is how Microsoft declare overloading methods in their projects, such as Language Server Protocol. However, it breaks method-signature-style. (To reproduce, click the typescript-eslint playgound link)

The second and third style is correct for method-signature-style, but it fails the type check when implementing those methods in a class. If I remove the overloading number | string signature, then it fails the type check when invoking method foo of the interface. (To reproduce, click the TS playgound link)

TypeScript Playground Link
typescript-eslint Playground

Therefore, I need some suggestions:
Should I just disable method-signature-style for those parts, or open an issue in typescript-eslint, or open an issue in TypeScript?

[yf-yang]

Let me try to summarize: the rule method-signature-style means, when I am declaring an interface method, it is possible that someone implements with either foo = () => {} or foo() {}. For the first one, method style may break typescript strict function check.

So, if I can guarantee this interface is only implemented in classes whose methods are declared in the second style, then I can safely ignore method-signature-style. Therefore, I should disable this rule in such scenario.

Is it correct?

[bradzacher]

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