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authorOndřej Bílka <neleai@seznam.cz>2013-10-30 16:07:15 +0100
committerOndřej Bílka <neleai@seznam.cz>2013-10-30 16:08:12 +0100
commitbbea82f7fe8af40fd08e8956e1aaf4d877168652 (patch)
treea10c1817228461fb3cd672a10754a78d6783e821 /stdlib/div.c
parent977f4b31b7ca4a4e498c397f3fd70510694bbd86 (diff)
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Remove code from div that is by C99 obsolete. Fixes bug 15799
Diffstat (limited to 'stdlib/div.c')
-rw-r--r--stdlib/div.c22
1 files changed, 0 insertions, 22 deletions
diff --git a/stdlib/div.c b/stdlib/div.c
index 44a30a7..0f5569a 100644
--- a/stdlib/div.c
+++ b/stdlib/div.c
@@ -59,27 +59,5 @@ div (numer, denom)
result.quot = numer / denom;
result.rem = numer % denom;
- /* The ANSI standard says that |QUOT| <= |NUMER / DENOM|, where
- NUMER / DENOM is to be computed in infinite precision. In
- other words, we should always truncate the quotient towards
- zero, never -infinity. Machine division and remainer may
- work either way when one or both of NUMER or DENOM is
- negative. If only one is negative and QUOT has been
- truncated towards -infinity, REM will have the same sign as
- DENOM and the opposite sign of NUMER; if both are negative
- and QUOT has been truncated towards -infinity, REM will be
- positive (will have the opposite sign of NUMER). These are
- considered `wrong'. If both are NUM and DENOM are positive,
- RESULT will always be positive. This all boils down to: if
- NUMER >= 0, but REM < 0, we got the wrong answer. In that
- case, to get the right answer, add 1 to QUOT and subtract
- DENOM from REM. */
-
- if (numer >= 0 && result.rem < 0)
- {
- ++result.quot;
- result.rem -= denom;
- }
-
return result;
}