170 lines
4.7 KiB
ArmAsm
170 lines
4.7 KiB
ArmAsm
/* $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
|