1 files changed, 0 insertions, 57 deletions
diff --git a/sysdeps/generic/lldiv.c b/sysdeps/generic/lldiv.c
deleted file mode 100644
index 28a016b744..0000000000
--- a/sysdeps/generic/lldiv.c
+++ /dev/null
@@ -1,57 +0,0 @@
-/* `long long int' divison with remainder.
-   Copyright (C) 1992, 1996, 1997 Free Software Foundation, Inc.
-   This file is part of the GNU C Library.
-
-   The GNU C Library is free software; you can redistribute it and/or
-   modify it under the terms of the GNU Lesser General Public
-   License as published by the Free Software Foundation; either
-   version 2.1 of the License, or (at your option) any later version.
-
-   The GNU C Library is distributed in the hope that it will be useful,
-   but WITHOUT ANY WARRANTY; without even the implied warranty of
-   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
-   Lesser General Public License for more details.
-
-   You should have received a copy of the GNU Lesser General Public
-   License along with the GNU C Library; if not, write to the Free
-   Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
-   02111-1307 USA.  */
-
-#include <stdlib.h>
-
-
-/* Return the `lldiv_t' representation of NUMER over DENOM.  */
-lldiv_t
-lldiv (numer, denom)
-     long long int numer;
-     long long int denom;
-{
-  lldiv_t result;
-
-  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;
-}