Comment by matklad

I assume you ment to write `assert(subexpression != undefined)`?

This is resilient parsing --- we are parsing source code with syntax errors, but still want to produce a best-effort syntax tree. Although expression is required by the grammar, the `expression` function might still return nothing if the user typed some garbage there instead of a valid expression.

However, even if we return nothing due to garbage, there are two possible behaviors:

* We can consume no tokens, making a guess that what looks like "garbage" from the perspective of expression parser is actually a start of next larger syntax construct:

``` function f() { let x = foo(1, let not_garbage = 92; } ```

In this example, it would be smart to _not_ consume `let` when parsing `foo(`'s arglist.

* Alternatively, we can consume some tokens, guessing that the user _meant_ to write an expression there

``` function f() { let x = foo(1, /); } ```

In the above example, it would be smart to skip over `/`.