1 ; RUN: opt < %s -instsimplify
3 ; The mul can be proved to always overflow (turning a negative value
4 ; into a positive one) and thus results in undefined behaviour. At
5 ; the same time we were deducing from the nsw flag that that mul could
6 ; be assumed to have a negative value (since if not it has an undefined
7 ; value, which can be taken to be negative). We were reporting the mul
8 ; as being both positive and negative, firing an assertion!
9 define i1 @test1(i32 %a) {
13 %2 = mul nsw i32 %1, 4
15 %4 = icmp ne i32 %3, 0