diff options
Diffstat (limited to 'stdlib')
| -rw-r--r-- | stdlib/div.c | 22 | ||||
| -rw-r--r-- | stdlib/ldiv.c | 22 | ||||
| -rw-r--r-- | stdlib/lldiv.c | 22 |
3 files changed, 0 insertions, 66 deletions
diff --git a/stdlib/div.c b/stdlib/div.c index 44a30a7ea4..0f5569a5dd 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; } 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; } diff --git a/stdlib/lldiv.c b/stdlib/lldiv.c index d1202bf9f9..0da1a6afc1 100644 --- a/stdlib/lldiv.c +++ b/stdlib/lldiv.c @@ -30,27 +30,5 @@ lldiv (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; } |