Linux-2.6.12-rc2v2.6.12-rc2

Initial git repository build. I'm not bothering with the full history,
even though we have it. We can create a separate "historical" git
archive of that later if we want to, and in the meantime it's about
3.2GB when imported into git - space that would just make the early
git days unnecessarily complicated, when we don't have a lot of good
infrastructure for it.

Let it rip!

1 files changed, 169 insertions, 0 deletions

diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S
new file mode 100644
index 000000000000..a784720a8a22
--- /dev/null
+++ b/arch/sparc/lib/umul.S
@@ -0,0 +1,169 @@
+/* $Id: umul.S,v 1.4 1996/09/30 02:22:39 davem Exp $
+ * umul.S:      This routine was taken from glibc-1.09 and is covered
+ *              by the GNU Library General Public License Version 2.
+ */
+
+
+/*
+ * Unsigned multiply.  Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
+ * upper 32 bits of the 64-bit product).
+ *
+ * This code optimizes short (less than 13-bit) multiplies.  Short
+ * multiplies require 25 instruction cycles, and long ones require
+ * 45 instruction cycles.
+ *
+ * On return, overflow has occurred (%o1 is not zero) if and only if
+ * the Z condition code is clear, allowing, e.g., the following:
+ *
+ *      call    .umul
+ *      nop
+ *      bnz     overflow        (or tnz)
+ */
+
+        .globl .umul
+.umul:
+        or      %o0, %o1, %o4
+        mov     %o0, %y         ! multiplier -> Y
+
+        andncc  %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
+        be      Lmul_shortway   ! if zero, can do it the short way
+         andcc  %g0, %g0, %o4   ! zero the partial product and clear N and V
+
+        /*
+         * Long multiply.  32 steps, followed by a final shift step.
+         */
+        mulscc  %o4, %o1, %o4   ! 1
+        mulscc  %o4, %o1, %o4   ! 2
+        mulscc  %o4, %o1, %o4   ! 3
+        mulscc  %o4, %o1, %o4   ! 4
+        mulscc  %o4, %o1, %o4   ! 5
+        mulscc  %o4, %o1, %o4   ! 6
+        mulscc  %o4, %o1, %o4   ! 7
+        mulscc  %o4, %o1, %o4   ! 8
+        mulscc  %o4, %o1, %o4   ! 9
+        mulscc  %o4, %o1, %o4   ! 10
+        mulscc  %o4, %o1, %o4   ! 11
+        mulscc  %o4, %o1, %o4   ! 12
+        mulscc  %o4, %o1, %o4   ! 13
+        mulscc  %o4, %o1, %o4   ! 14
+        mulscc  %o4, %o1, %o4   ! 15
+        mulscc  %o4, %o1, %o4   ! 16
+        mulscc  %o4, %o1, %o4   ! 17
+        mulscc  %o4, %o1, %o4   ! 18
+        mulscc  %o4, %o1, %o4   ! 19
+        mulscc  %o4, %o1, %o4   ! 20
+        mulscc  %o4, %o1, %o4   ! 21
+        mulscc  %o4, %o1, %o4   ! 22
+        mulscc  %o4, %o1, %o4   ! 23
+        mulscc  %o4, %o1, %o4   ! 24
+        mulscc  %o4, %o1, %o4   ! 25
+        mulscc  %o4, %o1, %o4   ! 26
+        mulscc  %o4, %o1, %o4   ! 27
+        mulscc  %o4, %o1, %o4   ! 28
+        mulscc  %o4, %o1, %o4   ! 29
+        mulscc  %o4, %o1, %o4   ! 30
+        mulscc  %o4, %o1, %o4   ! 31
+        mulscc  %o4, %o1, %o4   ! 32
+        mulscc  %o4, %g0, %o4   ! final shift
+
+
+        /*
+         * Normally, with the shift-and-add approach, if both numbers are
+         * positive you get the correct result.  With 32-bit two's-complement
+         * numbers, -x is represented as
+         *
+         *                x                 32
+         *      ( 2  -  ------ ) mod 2  *  2
+         *                 32
+         *                2
+         *
+         * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s,
+         * we can treat this as if the radix point were just to the left
+         * of the sign bit (multiply by 2^32), and get
+         *
+         *      -x  =  (2 - x) mod 2
+         *
+         * Then, ignoring the `mod 2's for convenience:
+         *
+         *   x *  y     = xy
+         *  -x *  y     = 2y - xy
+         *   x * -y     = 2x - xy
+         *  -x * -y     = 4 - 2x - 2y + xy
+         *
+         * For signed multiplies, we subtract (x << 32) from the partial
+         * product to fix this problem for negative multipliers (see mul.s).
+         * Because of the way the shift into the partial product is calculated
+         * (N xor V), this term is automatically removed for the multiplicand,
+         * so we don't have to adjust.
+         *
+         * But for unsigned multiplies, the high order bit wasn't a sign bit,
+         * and the correction is wrong.  So for unsigned multiplies where the
+         * high order bit is one, we end up with xy - (y << 32).  To fix it
+         * we add y << 32.
+         */
+#if 0
+        tst     %o1
+        bl,a    1f              ! if %o1 < 0 (high order bit = 1),
+         add    %o4, %o0, %o4   ! %o4 += %o0 (add y to upper half)
+
+1:
+        rd      %y, %o0         ! get lower half of product
+        retl
+         addcc  %o4, %g0, %o1   ! put upper half in place and set Z for %o1==0
+#else
+        /* Faster code from tege@sics.se.  */
+        sra     %o1, 31, %o2    ! make mask from sign bit
+        and     %o0, %o2, %o2   ! %o2 = 0 or %o0, depending on sign of %o1
+        rd      %y, %o0         ! get lower half of product
+        retl
+         addcc  %o4, %o2, %o1   ! add compensation and put upper half in place
+#endif
+
+Lmul_shortway:
+        /*
+         * Short multiply.  12 steps, followed by a final shift step.
+         * The resulting bits are off by 12 and (32-12) = 20 bit positions,
+         * but there is no problem with %o0 being negative (unlike above),
+         * and overflow is impossible (the answer is at most 24 bits long).
+         */
+        mulscc  %o4, %o1, %o4   ! 1
+        mulscc  %o4, %o1, %o4   ! 2
+        mulscc  %o4, %o1, %o4   ! 3
+        mulscc  %o4, %o1, %o4   ! 4
+        mulscc  %o4, %o1, %o4   ! 5
+        mulscc  %o4, %o1, %o4   ! 6
+        mulscc  %o4, %o1, %o4   ! 7
+        mulscc  %o4, %o1, %o4   ! 8
+        mulscc  %o4, %o1, %o4   ! 9
+        mulscc  %o4, %o1, %o4   ! 10
+        mulscc  %o4, %o1, %o4   ! 11
+        mulscc  %o4, %o1, %o4   ! 12
+        mulscc  %o4, %g0, %o4   ! final shift
+
+        /*
+         * %o4 has 20 of the bits that should be in the result; %y has
+         * the bottom 12 (as %y's top 12).  That is:
+         *
+         *        %o4               %y
+         * +----------------+----------------+
+         * | -12- |   -20-  | -12- |   -20-  |
+         * +------(---------+------)---------+
+         *         -----result-----
+         *
+         * The 12 bits of %o4 left of the `result' area are all zero;
+         * in fact, all top 20 bits of %o4 are zero.
+         */
+
+        rd      %y, %o5
+        sll     %o4, 12, %o0    ! shift middle bits left 12
+        srl     %o5, 20, %o5    ! shift low bits right 20
+        or      %o5, %o0, %o0
+        retl
+         addcc  %g0, %g0, %o1   ! %o1 = zero, and set Z
+
+        .globl  .umul_patch
+.umul_patch:
+        umul    %o0, %o1, %o0
+        retl
+         rd     %y, %o1
+        nop