root/arch/sparc/lib/umul.S

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   1 /* $Id: umul.S,v 1.2 1995/11/25 00:59:08 davem Exp $
   2  * umul.S:      This routine was taken from glibc-1.09 and is covered
   3  *              by the GNU Library General Public License Version 2.
   4  */
   5 
   6 
   7 /*
   8  * Unsigned multiply.  Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
   9  * upper 32 bits of the 64-bit product).
  10  *
  11  * This code optimizes short (less than 13-bit) multiplies.  Short
  12  * multiplies require 25 instruction cycles, and long ones require
  13  * 45 instruction cycles.
  14  *
  15  * On return, overflow has occurred (%o1 is not zero) if and only if
  16  * the Z condition code is clear, allowing, e.g., the following:
  17  *
  18  *      call    .umul
  19  *      nop
  20  *      bnz     overflow        (or tnz)
  21  */
  22 
  23         .globl .umul
  24 .umul:
  25         or      %o0, %o1, %o4
  26         mov     %o0, %y         ! multiplier -> Y
  27         andncc  %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
  28         be      Lmul_shortway   ! if zero, can do it the short way
  29         andcc   %g0, %g0, %o4   ! zero the partial product and clear N and V
  30 
  31         /*
  32          * Long multiply.  32 steps, followed by a final shift step.
  33          */
  34         mulscc  %o4, %o1, %o4   ! 1
  35         mulscc  %o4, %o1, %o4   ! 2
  36         mulscc  %o4, %o1, %o4   ! 3
  37         mulscc  %o4, %o1, %o4   ! 4
  38         mulscc  %o4, %o1, %o4   ! 5
  39         mulscc  %o4, %o1, %o4   ! 6
  40         mulscc  %o4, %o1, %o4   ! 7
  41         mulscc  %o4, %o1, %o4   ! 8
  42         mulscc  %o4, %o1, %o4   ! 9
  43         mulscc  %o4, %o1, %o4   ! 10
  44         mulscc  %o4, %o1, %o4   ! 11
  45         mulscc  %o4, %o1, %o4   ! 12
  46         mulscc  %o4, %o1, %o4   ! 13
  47         mulscc  %o4, %o1, %o4   ! 14
  48         mulscc  %o4, %o1, %o4   ! 15
  49         mulscc  %o4, %o1, %o4   ! 16
  50         mulscc  %o4, %o1, %o4   ! 17
  51         mulscc  %o4, %o1, %o4   ! 18
  52         mulscc  %o4, %o1, %o4   ! 19
  53         mulscc  %o4, %o1, %o4   ! 20
  54         mulscc  %o4, %o1, %o4   ! 21
  55         mulscc  %o4, %o1, %o4   ! 22
  56         mulscc  %o4, %o1, %o4   ! 23
  57         mulscc  %o4, %o1, %o4   ! 24
  58         mulscc  %o4, %o1, %o4   ! 25
  59         mulscc  %o4, %o1, %o4   ! 26
  60         mulscc  %o4, %o1, %o4   ! 27
  61         mulscc  %o4, %o1, %o4   ! 28
  62         mulscc  %o4, %o1, %o4   ! 29
  63         mulscc  %o4, %o1, %o4   ! 30
  64         mulscc  %o4, %o1, %o4   ! 31
  65         mulscc  %o4, %o1, %o4   ! 32
  66         mulscc  %o4, %g0, %o4   ! final shift
  67 
  68 
  69         /*
  70          * Normally, with the shift-and-add approach, if both numbers are
  71          * positive you get the correct result.  With 32-bit two's-complement
  72          * numbers, -x is represented as
  73          *
  74          *                x                 32
  75          *      ( 2  -  ------ ) mod 2  *  2
  76          *                 32
  77          *                2
  78          *
  79          * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s,
  80          * we can treat this as if the radix point were just to the left
  81          * of the sign bit (multiply by 2^32), and get
  82          *
  83          *      -x  =  (2 - x) mod 2
  84          *
  85          * Then, ignoring the `mod 2's for convenience:
  86          *
  87          *   x *  y     = xy
  88          *  -x *  y     = 2y - xy
  89          *   x * -y     = 2x - xy
  90          *  -x * -y     = 4 - 2x - 2y + xy
  91          *
  92          * For signed multiplies, we subtract (x << 32) from the partial
  93          * product to fix this problem for negative multipliers (see mul.s).
  94          * Because of the way the shift into the partial product is calculated
  95          * (N xor V), this term is automatically removed for the multiplicand,
  96          * so we don't have to adjust.
  97          *
  98          * But for unsigned multiplies, the high order bit wasn't a sign bit,
  99          * and the correction is wrong.  So for unsigned multiplies where the
 100          * high order bit is one, we end up with xy - (y << 32).  To fix it
 101          * we add y << 32.
 102          */
 103 #if 0
 104         tst     %o1
 105         bl,a    1f              ! if %o1 < 0 (high order bit = 1),
 106         add     %o4, %o0, %o4   ! %o4 += %o0 (add y to upper half)
 107 1:      rd      %y, %o0         ! get lower half of product
 108         retl
 109         addcc   %o4, %g0, %o1   ! put upper half in place and set Z for %o1==0
 110 #else
 111         /* Faster code from tege@sics.se.  */
 112         sra     %o1, 31, %o2    ! make mask from sign bit
 113         and     %o0, %o2, %o2   ! %o2 = 0 or %o0, depending on sign of %o1
 114         rd      %y, %o0         ! get lower half of product
 115         retl
 116         addcc   %o4, %o2, %o1   ! add compensation and put upper half in place
 117 #endif
 118 
 119 Lmul_shortway:
 120         /*
 121          * Short multiply.  12 steps, followed by a final shift step.
 122          * The resulting bits are off by 12 and (32-12) = 20 bit positions,
 123          * but there is no problem with %o0 being negative (unlike above),
 124          * and overflow is impossible (the answer is at most 24 bits long).
 125          */
 126         mulscc  %o4, %o1, %o4   ! 1
 127         mulscc  %o4, %o1, %o4   ! 2
 128         mulscc  %o4, %o1, %o4   ! 3
 129         mulscc  %o4, %o1, %o4   ! 4
 130         mulscc  %o4, %o1, %o4   ! 5
 131         mulscc  %o4, %o1, %o4   ! 6
 132         mulscc  %o4, %o1, %o4   ! 7
 133         mulscc  %o4, %o1, %o4   ! 8
 134         mulscc  %o4, %o1, %o4   ! 9
 135         mulscc  %o4, %o1, %o4   ! 10
 136         mulscc  %o4, %o1, %o4   ! 11
 137         mulscc  %o4, %o1, %o4   ! 12
 138         mulscc  %o4, %g0, %o4   ! final shift
 139 
 140         /*
 141          * %o4 has 20 of the bits that should be in the result; %y has
 142          * the bottom 12 (as %y's top 12).  That is:
 143          *
 144          *        %o4               %y
 145          * +----------------+----------------+
 146          * | -12- |   -20-  | -12- |   -20-  |
 147          * +------(---------+------)---------+
 148          *         -----result-----
 149          *
 150          * The 12 bits of %o4 left of the `result' area are all zero;
 151          * in fact, all top 20 bits of %o4 are zero.
 152          */
 153 
 154         rd      %y, %o5
 155         sll     %o4, 12, %o0    ! shift middle bits left 12
 156         srl     %o5, 20, %o5    ! shift low bits right 20
 157         or      %o5, %o0, %o0
 158         retl
 159         addcc   %g0, %g0, %o1   ! %o1 = zero, and set Z

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