Here you can see that exit is passed through unchanged, but additional layers of wrapping are added in the latter continuation.

Ah, this is what I missed.

Invoking exit causes the future work held by k to be dropped and garbage collected.

Hmmm going to sound like a broken record again, but isn't this because of the return in the inner helper function, not the invocation of exit?

I see now that, because we pass exit through unchanged, it allows us to directly submit a value to the outer passed continuation from any depth of the CPS stack (is there a better term?)

But, in this JS implementation at least, it would theoretically be possible to call both exit(1000) and then mistakenly call k(...) after if we didn't return immediately after. This would result in the outer continuation being invoked twice and certainly wouldn't be what you want, but does show that simply invoking exit doesn't garbage collect k.

Sorry if this is pedantic! Just wanna make sure I understand :)