On Fri, Jan 4, 2019 at 3:23 PM Jonas Devlieghere <jo...@devlieghere.com> wrote:
> On Fri, Jan 4, 2019 at 3:13 PM Zachary Turner <ztur...@google.com> wrote: > >> I don't think #2 is a correct change. Just because the sign bit is set >> doesn't mean it's signed. Is the 4-byte value 0x10000000 signed or >> unsigned? It's a trick question, because there's not enough information! >> If it was written "int x = 0x10000000" then it's signed (and negative). If >> it was written "unsigned x = 0x10000000" then it's unsigned (and >> positive). What about the 4-byte value 0x1? Still a trick! If it was >> written "int x = 1" then it's signed (and positive), and if it was written >> "unsigned x = 1" then it's unsigned (and positive). >> >> My point is that signedness of the *type* does not necessarly imply >> signedness of the value, and vice versa. >> >> APInt is purely a bit-representation and a size, there is no information >> whatsoever about whether type *type* is signed. It doesn't make sense to >> say "is this APInt negative?" without additional information. >> >> With APSInt, on the other hand, it does make sense to ask that question. >> If you have an APSInt where isSigned() is true, *then* you can use the sign >> bit to determine whether it's negative. And if you have an APSInt where >> isSigned() is false, then the "sign bit" is not actually a sign bit at all, >> it is just an extra power of 2 for the unsigned value. >> >> This is my understanding of the classes, someone correct me if I'm wrong. >> > >> IIUC though, the way to fix this is by using APSInt throughout the class, >> and delete all references to APInt. >> > > I think we share the same understanding. If we know at every call site > whether the type is signed or not then I totally agree, we should only use > APSInt. The reason I propose doing (2) first is for the first scenario you > described, where you don't know. Turning it into an explicit APSInt is as > bad as using an APInt and looking at the value. The latter has the > advantage that it conveys that you don't know, while the other may or may > not be a lie. > Do we ever not know though? And if so, then why don't we know whether the type is supposed to be signed or unsigned? Because guessing is always going to be wrong sometimes.
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