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/*
* 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)
*/
#include "DEFS.h"
FUNC(.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
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