Add qemu 2.4.0

Change-Id: Ic99cbad4b61f8b127b7dc74d04576c0bcbaaf4f5
Signed-off-by: Yang Zhang <yang.z.zhang@intel.com>

1 files changed, 43 insertions, 0 deletions

diff --git a/qemu/roms/u-boot/lib/ldiv.c b/qemu/roms/u-boot/lib/ldiv.c
new file mode 100644
index 000000000..8e11333ac
--- /dev/null
+++ b/qemu/roms/u-boot/lib/ldiv.c
@@ -0,0 +1,43 @@
+/* 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;
+}