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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 <sys/syscall.h>


.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