This is a continuation of the series of patches adding lane wise support for scalable vectors in various knownbit-esq routines.
The basic idea here is that we track a single lane for scalable vectors which corresponds to an unknown number of lanes at runtime. This is enough for us to perform lane wise reasoning on many arithmetic operations.
This can never be true can it?
More relevant to the other exit paths but is return 1; what we want here rather than break;? I asked because the existing scalable vector early exits do the latter (e.g. a couple of lines down) and so I'm wondering if we're loosing information or if the other exit paths are wrong.