[zarzavat]

It can’t strip what’s after the as keyword without an up-to-date TS grammar, because `as` is an expression. The parser needs to know how to parse type expressions in order to know when the RHS of the `as` expression ends.

Let’s say that typescript adds a new type operator “wobble T”. What does this desugar to?

    x as wobble
    T

Without knowing about the new wobble syntax this would be parsed as `x as wobble; T` and desugar to `x; T`

With the new wobble syntax it would be parsed as `x as (wobble T);` according to JS semicolon insertion rules because the expression wobble is incomplete, and desugar to `x`

[bazoom42]

The “as” expression is not valid JavaScript anyway, so the default rule for implicit semicolon does not apply. A grammer for type expressions could define if and how semicolons should be inserted.

[zarzavat]

TypeScript already has such type operators though. For example:

    type T = keyof
    {
      a: null,
      b: null
    }

Here T is “a”|”b”, no automatic semicolon is inserted after `keyof`. While I don’t personally write code like this, I’m sure that someone does. It’s perfectly within the rules, after all.

While it’s true that TS doesn’t have to follow JS rules for semicolon insertion in type expressions, it always has done, and probably always should do.

[bazoom42]

This is just the default. Automatic semicolon insertion only happen in specific well-defined cases, for example after the “return” keyword or when an invalid expression can be made valid by semicolon insertion. Neither applies here.