Hi,
Since bug 30255 has been declared as never going to be fixed, I've been
enjoying going through half a million lines of code looking for places where I
have to declare things long double to keep gcc from arbitrarily rounding down
intermediate results. The problem now is that I have come across a case where
the exact opposite occurs: If I declare a variable long double, gcc inserts a
round to double before computing the square root, where it does not if I
declare it double. I will upload the file seperately, but here's the section
of code:
for (i=0; i < N; i++) t0 += X[i]*X[i];
t0 = sqrt(t0);
When compiled with t0 declared as double, no spills are performed by gcc, and
80-bit accuracy is maintained throughout the computation (critical to avoid
overflow). When declared as long double, however, the following code is
inserted:
fstpl 16(%rsp)
fldl 16(%rsp)
fld %st(0)
fsqrt
So, a long double is rounded to double by gcc, even though there is no store in
the algorithm. Any idea what is going on, and is there anything to be done? I
will post the short file seperately. You can gen both assemblies to see the
difference with
gcc -O -mfpmath=387 -S nrm2.c # gen double declaration
gcc -O -mfpmath=387 -DLD_ -S nrm2.c # gen long double variant
Thanks,
Clint
--
Summary: long double declaration rounds to double instead
Product: gcc
Version: 4.1.1
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: target
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: whaley at cs dot utsa dot edu
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=30599
