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[Bug libc/14551] New: strtold overflow handling for IBM long double
- From: "jsm28 at gcc dot gnu.org" <sourceware-bugzilla at sourceware dot org>
- To: glibc-bugs at sources dot redhat dot com
- Date: Thu, 06 Sep 2012 15:57:13 +0000
- Subject: [Bug libc/14551] New: strtold overflow handling for IBM long double
- Auto-submitted: auto-generated
http://sourceware.org/bugzilla/show_bug.cgi?id=14551
Bug #: 14551
Summary: strtold overflow handling for IBM long double
Product: glibc
Version: 2.16
Status: NEW
Severity: normal
Priority: P2
Component: libc
AssignedTo: unassigned@sourceware.org
ReportedBy: jsm28@gcc.gnu.org
CC: drepper.fsp@gmail.com
Classification: Unclassified
Host: powerpc*-linux*
The core strtold code treats IBM long double as an IEEE type with a fixed
mantissa precision of 106 bits. It's then
sysdeps/ieee754/ldbl-128ibm/mpn2ldbl.c that actually encodes the result in long
double.
Where the value overflows for IBM long double (because of the requirement that
the two parts, added together in round-to-nearest mode, equal the top part of
the long double) but not for the 106-bit IEEE type, the mpn2ldbl code generates
an invalid long double value. It should generate the appropriate overflowed
result (typically an infinity, but properly depending on rounding mode), set
errno to ERANGE and raise the overflow exception. Testcase:
#include <stdio.h>
#include <stdlib.h>
int
main (void)
{
union { long double ld; double d[2]; } x;
x.ld = strtold ("0x1.fffffffffffff8p+1023", NULL);
printf ("%a %a\n", x.d[0], x.d[1]);
return 0;
}
This prints:
inf -0x1p+970
That's not a valid infinity (and so causes problems such as not comparing equal
to a proper infinity); the least significant part of an infinity must be +0 or
-0 as per GCC's libgcc/config/rs6000/ibm-ldouble-format.
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