root/arch/sparc/lib/umul.S

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

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