FunctionLike = SignatureDeclaration

[ghost]

FunctionLike and SignatureDeclaration wind up being identical types once all sub-unions are flattened. You can test this out with:

import ts = require("typescript");
let a: ts.FunctionLike = <any>null;
let b: ts.SignatureDeclaration = <any>null;
a = b;
b = a;

My suggested fix would be to just use SignatureDeclaration everywhere.

(Ref: #16988)

[weswigham]

I am willing to believe that this is currently the case, even though it probably wasn't always so 😛

[DanielRosenwasser]

Maybe the appropriate change is to keep FunctionLike and brand the constituents instead?

[ghost]

Summary of offline discussion with @DanielRosenwasser: We don't need a brand type because SignatureDeclaration is already a union type, which supports narrowing already.