Linux-2.6.12-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!
diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S
new file mode 100644
index 0000000..a784720
--- /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