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...} } ==> (step 2) known bits analysis on sext to prove divisor is non-zero i32 prod = x *nsw y i64 prod.sext = sext prod to i64 i64 t = K `udiv` (-1 + (prod.sext >> 32)) for (...) { if (condition){ use(t) } } ==> (step 3) commute sext through nsw multiplicationa ;; i32 prod = x *nsw y i64 prod.sext = (sext x) *nsw (sext y) i64 t = K `udiv` (-1 + (prod.sext >> 32)) for (...) { if (condition){ use(t) } } Now we've managed to introduce a fault if we were in case 1 -- the source program was well defined but the targ...

...-1 + (prod.sext >> 32)) use(t) } } ==> (step 2) known bits analysis on sext to prove divisor is non-zero i32 prod = x *nsw y i64 prod.sext = sext prod to i64 i64 t = K `udiv` (-1 + (prod.sext >> 32)) for (...) { if (condition){ use(t) } } ==> (step 3) commute sext through nsw multiplicationa ;; i32 prod = x *nsw y i64 prod.sext = (sext x) *nsw (sext y) i64 t = K `udiv` (-1 + (prod.sext >> 32)) for (...) { if (condition){ use(t) } } Now we've managed to introduce a fault if we were in case 1 -- the source program was well defined but the target program divides by zero. _...

Hi John, John McCall wrote: >> On Nov 10, 2016, at 10:37 PM, Sanjoy Das<sanjoy at playingwithpointers.com> wrote: >> As a concrete example, say we have: >> >> i32 sum = x *nsw y >> i64 sum.sext = sext sum to i64 >> ... >> some use of sum.sext >> >> >> Pretending "x +nsw 1" does not sign-overflow, we can commute the sext