int triangle(int a,int b) { int c; c=a*b/2; return c; }
emits this very bizarre code (at -O, -O2):
mov a,%edx
mov b,%eax
imull %edx,%eax
movl %eax,%edx
shrl $31,%edx
addl %edx,%eax
sarl %eax
ret
Why are the two instructions after the imull emitted? Shouldn't this become
simply imull and sarl? This code extracts the most significant bit of a*b and
adds it to a*b, then shifting the result right. It almost looks as if trying to
round or something? There is probably something obvious I'm overlooking here.
The analog code is generated for ppc, x86_64, sparc and mips. Please explain.
I also tried with -Os, and the code becomes cltd (sign-extend 32 to 64 bits)
plus idivl with 2. Could it be that the peephole optimizer converts the idivl
to a shift but forgets to remove the sign extend code?
--
Summary: bizarre code for int*int/2
Product: gcc
Version: 3.4.4
Status: UNCONFIRMED
Severity: minor
Priority: P2
Component: c
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: felix-gcc at fefe dot de
CC: gcc-bugs at gcc dot gnu dot org
GCC build triplet: i686-pc-linux-gnu
GCC host triplet: i686-pc-linux-gnu
GCC target triplet: i686-pc-linux-gnu
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=22072
