How to cast object to another type and remove unneeded fields in TypeScript?

Question

Suppose I have the following code:

interface A {     a: number; }  interface B extends A {     b: number; }  const b: B = {a: 1, b: 5}; const a: A = b as A; 

Now the variable a has type A, but it still contains b inside of it. Sometimes it is undesirable - I'd like to be sure that, if I have a variable of type A, it has the exact fields of type A. I was wondering, whether TypeScript has some kind of a "hard cast" which would remove any unneeded fields when converting between types.

Answer 1

There is no casting in TypeScript because that's not how TypeScript works. TS is a type layer on top of JS. When you assign a type in TS you don't set a type, you annotate that the given expression is of a certain type. When the TS code is transpiled to JS, all type information is stripped.

Or in other words: the type system is JS, with TS you just announce that a variable is of a certain type. More often than never it happens that you assign the wrong type at design time and get suprised during debug that the variable has a completely different type than expected.

If you want to be sure that a property is removed from an object, you need to go the JS way (and - if necessary - annotate the result with TS). Check this Q&A to see how to remove properties from an object.