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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 

.global   .umul;

.align 4;

.type  .umul ,@function;

.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; clear N & 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

.size  .umul , . -.umul