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author | Ondřej Bílka <neleai@seznam.cz> | 2013-10-30 16:07:15 +0100 |
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committer | Ondřej Bílka <neleai@seznam.cz> | 2013-10-30 16:08:12 +0100 |
commit | bbea82f7fe8af40fd08e8956e1aaf4d877168652 (patch) | |
tree | a10c1817228461fb3cd672a10754a78d6783e821 /stdlib/ldiv.c | |
parent | 977f4b31b7ca4a4e498c397f3fd70510694bbd86 (diff) | |
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Remove code from div that is by C99 obsolete. Fixes bug 15799
Diffstat (limited to 'stdlib/ldiv.c')
-rw-r--r-- | stdlib/ldiv.c | 22 |
1 files changed, 0 insertions, 22 deletions
diff --git a/stdlib/ldiv.c b/stdlib/ldiv.c index 76d474fc62..a03057fc0d 100644 --- a/stdlib/ldiv.c +++ b/stdlib/ldiv.c @@ -27,27 +27,5 @@ ldiv (long int numer, long int 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; } |