https://gcc.gnu.org/bugzilla/show_bug.cgi?id=101703
Bug ID: 101703
Summary: (bool0 + bool1) & 1 and (bool0 + bool1) == 1 can be
optimized to bool0 ^ bool1
Product: gcc
Version: 12.0
Status: UNCONFIRMED
Keywords: missed-optimization
Severity: enhancement
Priority: P3
Component: tree-optimization
Assignee: unassigned at gcc dot gnu.org
Reporter: pinskia at gcc dot gnu.org
Target Milestone: ---
Take:
bool f(bool a, bool b)
{
int t = a;
int t1 = b;
return (t + t1) & 1;
}
bool fa(bool a, bool b)
{
int t = a;
int t1 = b;
return (t + t1)==1;
}
bool fb(bool a, bool b)
{
return a!=b;
}
bool fc(bool a, bool b)
{
return a^b;
}
These three should produce the same code gen. Right now fb and fc do but f and
fa needs to handled.
the for fa, == 1 can be converted into & 1 as the range is [0,2]:
# RANGE [0, 2] NONZERO 3
_1 = t_3 + t1_5;
_6 = _1 == 1;
and only 1 can be if & 1 is true. And the rest just follows through.
