sparc32: Kill off software 32-bit multiply/divide routines.

For the explicit calls to .udiv/.umul in assembler, I made a
mechanical (read as: safe) transformation.  I didn't attempt
to make any simplifications.

In particular, __ndelay and __udelay can be simplified significantly.
Some of the %y reads are unnecessary and these routines have no need
any longer for allocating a register window, they can be leaf
functions.

Signed-off-by: David S. Miller <davem@davemloft.net>

1 files changed, 0 insertions, 171 deletions

diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S
deleted file mode 100644
index 1f36ae682529..000000000000
--- a/arch/sparc/lib/umul.S
+++ /dev/null
@@ -1,171 +0,0 @@
-/*
- * 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
-	.globl _Umul
-.umul:
-_Umul:	/* needed for export */
-	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