Alright, let's dive into how null is handled in Typescript.
In Typescript, the treatment of null and undefined differs based on whether the --strictNullChecks flag is enabled. When true, null and undefined are treated as literal types with only one member each - null and undefined, respectively. However, when false, null and undefined are considered members of every type.
Now, let's discuss intersection types. For example, type T = A & B means that a value must belong to both type A and type B to be classified as type T. Therefore, {a:string} & {b:string} results in {a:string, b:string}. Intersections involving non-interface types can lead to nonsensical combinations. What would string & number represent? Since no value can be both a string and a number simultaneously, this type has no valid instances.
Moving on to your code snippet, defining type TypeB = null essentially acts as an alias for the null type. Consequently, trying to assign an object like {id: '1'} to a variable declared as TypeB will fail to compile in Typescript since an object cannot be null.
In another scenario where
type TypeB = InterfaceA & null
, the behavior depends on compiler settings. With strict null checks active, TypeB becomes InterfaceA. Conversely, with these checks disabled, it defaults to null. It may seem puzzling at first, but ultimately, this construct serves no practical purpose. The compilation success seen is often due to the use of
as TypeB, which hints at unsafe type assertions. These should ideally be limited to cases involving unknown data sources, such as deserialized information.
To sum up, the intention behind the given code remains unclear. Consider reviewing the official Typescript documentation to gain a deeper understanding and clarity on how to proceed.