On Tue, 2011-05-24 at 09:18 -0700, H. Peter Anvin wrote: I'll do it that way if you want. But you now have an extra jb and an extra movl, neither of which do anything at all. It's no different than movq %rax, %rsi cmp{q,l} $-MAX_ERRNO, %{r,e}ax setae %al I know it's the same because I spent forever trying to hunt down movslq. I don't understand why it's not in the Intel® 64 and IA-32 Architectures Software Developer’s Manual Volume 2 (2A & 2B): Instruction Set Reference, A-Z. That's exactly what I talked about, truncating the upper 32 bits just the sign extend them right back. I guess it comes down to picking one of these 3: My version: movq %rax,%rsi /* second arg, syscall return value */ cmpl $-MAX_ERRNO,%rax /* is it < 0? */ setbe %al /* 1 if so, 0 if not */ movzbl %al,%edi /* zero-extend that into %edi */ call __audit_syscall_exit VS hpa version: movl %eax,%esi /* move 32bits to second arg */ cmpl $-MAX_ERRNO,%eax /* check if fail */ jbe 1f movslq %eax, %rsi /* re-sign-extend eax */ 1: setbe %al movzbl %al,%edi call __audit_syscall_exit VS alternate of hpa version without set: movl $1,%edi /* syscall success argument */ movl %eax,%esi /* move 32bits to second arg */ cmpl $-MAX_ERRNO,%eax /* check if fail */ jbe 1f xor %edi,%edi /* syscall failure argument */ movslq %eax, %rsi /* resign-extend eax */ 1: call __audit_syscall_exit If I have to go with the hpa version of truncation followed by sign extension, is it any better/cheap/faster to use just movl in the 'common' case and movl+xor in the uncommon syscall failure? I don't know how expensive or large the set+movzbl is....