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-rw-r--r--qemu/roms/u-boot/lib/ldiv.c43
1 files changed, 0 insertions, 43 deletions
diff --git a/qemu/roms/u-boot/lib/ldiv.c b/qemu/roms/u-boot/lib/ldiv.c
deleted file mode 100644
index 8e11333ac..000000000
--- a/qemu/roms/u-boot/lib/ldiv.c
+++ /dev/null
@@ -1,43 +0,0 @@
-/* Copyright (C) 1992, 1997 Free Software Foundation, Inc.
- This file is part of the GNU C Library.
-
- * SPDX-License-Identifier: LGPL-2.0+
- */
-
-typedef struct {
- long quot;
- long rem;
-} ldiv_t;
-/* Return the `ldiv_t' representation of NUMER over DENOM. */
-ldiv_t
-ldiv (long int numer, long int denom)
-{
- ldiv_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;
-}