/* This is an assembly language implementation of mulsi3, divsi3, and modsi3
   for the sparc processor.

   These routines are derived from the SPARC Architecture Manual, version 8,
   slightly edited to match the desired calling convention, and also to
   optimize them for our purposes.  */

#ifdef L_mulsi3
.text
        .align 4
        .global .umul
        .proc 4
.umul:
        or        %o0, %o1, %o4        ! logical or of multiplier and multiplicand
        mov        %o0, %y                ! multiplier to Y register
        andncc        %o4, 0xfff, %o5        ! mask out lower 12 bits
        be        mul_shortway        ! can do it the short way
        andcc        %g0, %g0, %o4        ! zero the partial product and clear NV cc
        !
        ! long multiply
        !
        mulscc        %o4, %o1, %o4        ! first iteration of 33
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4        ! 32nd iteration
        mulscc        %o4, %g0, %o4        ! last iteration only shifts
        ! the upper 32 bits of product are wrong, but we do not care
        retl
        rd        %y, %o0
        !
        ! short multiply
        !
mul_shortway:
        mulscc        %o4, %o1, %o4        ! first iteration of 13
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4
        mulscc        %o4, %o1, %o4        ! 12th iteration
        mulscc        %o4, %g0, %o4        ! last iteration only shifts
        rd        %y, %o5
        sll        %o4, 12, %o4        ! left shift partial product by 12 bits
        srl        %o5, 20, %o5        ! right shift partial product by 20 bits
        retl
        or        %o5, %o4, %o0        ! merge for true product
#endif

#ifdef L_divsi3
/*
 * Division and remainder, from Appendix E of the SPARC Version 8
 * Architecture Manual, with fixes from Gordon Irlam.
 */

/*
 * Input: dividend and divisor in %o0 and %o1 respectively.
 *
 * m4 parameters:
 *  .div        name of function to generate
 *  div                div=div => %o0 / %o1; div=rem => %o0 % %o1
 *  true                true=true => signed; true=false => unsigned
 *
 * Algorithm parameters:
 *  N                how many bits per iteration we try to get (4)
 *  WORDSIZE        total number of bits (32)
 *
 * Derived constants:
 *  TOPBITS        number of bits in the top decade of a number
 *
 * Important variables:
 *  Q                the partial quotient under development (initially 0)
 *  R                the remainder so far, initially the dividend
 *  ITER        number of main division loop iterations required;
 *                equal to ceil(log2(quotient) / N).  Note that this
 *                is the log base (2^N) of the quotient.
 *  V                the current comparand, initially divisor*2^(ITER*N-1)
 *
 * Cost:
 *  Current estimate for non-large dividend is
 *        ceil(log2(quotient) / N) * (10 + 7N/2) + C
 *  A large dividend is one greater than 2^(31-TOPBITS) and takes a
 *  different path, as the upper bits of the quotient must be developed
 *  one bit at a time.
 */
        .global .udiv
        .align 4
        .proc 4
        .text
.udiv:
         b ready_to_divide
         mov 0, %g3             ! result is always positive

        .global .div
        .align 4
        .proc 4
        .text
.div:
        ! compute sign of result; if neither is negative, no problem
        orcc        %o1, %o0, %g0        ! either negative?
        bge        ready_to_divide        ! no, go do the divide
        xor        %o1, %o0, %g3        ! compute sign in any case
        tst        %o1
        bge        1f
        tst        %o0
        ! %o1 is definitely negative; %o0 might also be negative
        bge        ready_to_divide        ! if %o0 not negative...
        sub        %g0, %o1, %o1        ! in any case, make %o1 nonneg
1:        ! %o0 is negative, %o1 is nonnegative
        sub        %g0, %o0, %o0        ! make %o0 nonnegative


ready_to_divide:

        ! Ready to divide.  Compute size of quotient; scale comparand.
        orcc        %o1, %g0, %o5
        bne        1f
        mov        %o0, %o3

        ! Divide by zero trap.  If it returns, return 0 (about as
        ! wrong as possible, but that is what SunOS does...).
        ta        0x2                    ! ST_DIV0
        retl
        clr        %o0

1:
        cmp        %o3, %o5                ! if %o1 exceeds %o0, done
        blu        got_result                ! (and algorithm fails otherwise)
        clr        %o2
        sethi        %hi(1 << (32 - 4 - 1)), %g1
        cmp        %o3, %g1
        blu        not_really_big
        clr        %o4

        ! Here the dividend is >= 2**(31-N) or so.  We must be careful here,
        ! as our usual N-at-a-shot divide step will cause overflow and havoc.
        ! The number of bits in the result here is N*ITER+SC, where SC <= N.
        ! Compute ITER in an unorthodox manner: know we need to shift V into
        ! the top decade: so do not even bother to compare to R.
        1:
                cmp        %o5, %g1
                bgeu        3f
                mov        1, %g2
                sll        %o5, 4, %o5
                b        1b
                add        %o4, 1, %o4

        ! Now compute %g2.
        2:        addcc        %o5, %o5, %o5
                bcc        not_too_big
                add        %g2, 1, %g2

                ! We get here if the %o1 overflowed while shifting.
                ! This means that %o3 has the high-order bit set.
                ! Restore %o5 and subtract from %o3.
                sll        %g1, 4, %g1        ! high order bit
                srl        %o5, 1, %o5        ! rest of %o5
                add        %o5, %g1, %o5
                b        do_single_div
                sub        %g2, 1, %g2

        not_too_big:
        3:        cmp        %o5, %o3
                blu        2b
                nop
                be        do_single_div
                nop
        /* NB: these are commented out in the V8-SPARC manual as well */
        /* (I do not understand this) */
        ! %o5 > %o3: went too far: back up 1 step
        !        srl        %o5, 1, %o5
        !        dec        %g2
        ! do single-bit divide steps
        !
        ! We have to be careful here.  We know that %o3 >= %o5, so we can do the
        ! first divide step without thinking.  BUT, the others are conditional,
        ! and are only done if %o3 >= 0.  Because both %o3 and %o5 may have the high-
        ! order bit set in the first step, just falling into the regular
        ! division loop will mess up the first time around.
        ! So we unroll slightly...
        do_single_div:
                subcc        %g2, 1, %g2
                bl        end_regular_divide
                nop
                sub        %o3, %o5, %o3
                mov        1, %o2
                b        end_single_divloop
                nop
        single_divloop:
                sll        %o2, 1, %o2
                bl        1f
                srl        %o5, 1, %o5
                ! %o3 >= 0
                sub        %o3, %o5, %o3
                b        2f
                add        %o2, 1, %o2
        1:        ! %o3 < 0
                add        %o3, %o5, %o3
                sub        %o2, 1, %o2
        2:
        end_single_divloop:
                subcc        %g2, 1, %g2
                bge        single_divloop
                tst        %o3
                b,a        end_regular_divide

not_really_big:
1:
        sll        %o5, 4, %o5
        cmp        %o5, %o3
        bleu        1b
        addcc        %o4, 1, %o4
        be        got_result
        sub        %o4, 1, %o4

        tst        %o3        ! set up for initial iteration
divloop:
        sll        %o2, 4, %o2
        ! depth 1, accumulated bits 0
        bl        L1.16
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 2, accumulated bits 1
        bl        L2.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 3, accumulated bits 3
        bl        L3.19
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 7
        bl        L4.23
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (7*2+1), %o2
        
L4.23:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (7*2-1), %o2
        
        
L3.19:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 5
        bl        L4.21
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (5*2+1), %o2
        
L4.21:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (5*2-1), %o2
        
L2.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 3, accumulated bits 1
        bl        L3.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 3
        bl        L4.19
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (3*2+1), %o2
        
L4.19:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (3*2-1), %o2

L3.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 1
        bl        L4.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (1*2+1), %o2

L4.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (1*2-1), %o2
        
L1.16:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 2, accumulated bits -1
        bl        L2.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 3, accumulated bits -1
        bl        L3.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -1
        bl        L4.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-1*2+1), %o2
        
L4.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-1*2-1), %o2
        
L3.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -3
        bl        L4.13
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-3*2+1), %o2
        
L4.13:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-3*2-1), %o2
        
L2.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 3, accumulated bits -3
        bl        L3.13
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -5
        bl        L4.11
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-5*2+1), %o2
        
L4.11:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-5*2-1), %o2
        
L3.13:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -7
        bl        L4.9
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-7*2+1), %o2

L4.9:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-7*2-1), %o2
        
        9:
end_regular_divide:
        subcc        %o4, 1, %o4
        bge        divloop
        tst        %o3
        bl,a        got_result
        ! non-restoring fixup here (one instruction only!)
        sub        %o2, 1, %o2


got_result:
        ! check to see if answer should be < 0
        tst        %g3
        bl,a        1f
        sub %g0, %o2, %o2
1:
        retl
        mov %o2, %o0
#endif

#ifdef L_modsi3
/* This implementation was taken from glibc:
 *
 * Input: dividend and divisor in %o0 and %o1 respectively.
 *
 * Algorithm parameters:
 *  N                how many bits per iteration we try to get (4)
 *  WORDSIZE        total number of bits (32)
 *
 * Derived constants:
 *  TOPBITS        number of bits in the top decade of a number
 *
 * Important variables:
 *  Q                the partial quotient under development (initially 0)
 *  R                the remainder so far, initially the dividend
 *  ITER        number of main division loop iterations required;
 *                equal to ceil(log2(quotient) / N).  Note that this
 *                is the log base (2^N) of the quotient.
 *  V                the current comparand, initially divisor*2^(ITER*N-1)
 *
 * Cost:
 *  Current estimate for non-large dividend is
 *        ceil(log2(quotient) / N) * (10 + 7N/2) + C
 *  A large dividend is one greater than 2^(31-TOPBITS) and takes a
 *  different path, as the upper bits of the quotient must be developed
 *  one bit at a time.
 */
.text
        .align 4
        .global        .urem
        .proc 4
.urem:
        b        divide
        mov        0, %g3                ! result always positive

        .align 4
        .global .rem
        .proc 4
.rem:
        ! compute sign of result; if neither is negative, no problem
        orcc        %o1, %o0, %g0        ! either negative?
        bge        2f                        ! no, go do the divide
        mov        %o0, %g3                ! sign of remainder matches %o0
        tst        %o1
        bge        1f
        tst        %o0
        ! %o1 is definitely negative; %o0 might also be negative
        bge        2f                        ! if %o0 not negative...
        sub        %g0, %o1, %o1        ! in any case, make %o1 nonneg
1:        ! %o0 is negative, %o1 is nonnegative
        sub        %g0, %o0, %o0        ! make %o0 nonnegative
2:

        ! Ready to divide.  Compute size of quotient; scale comparand.
divide:
        orcc        %o1, %g0, %o5
        bne        1f
        mov        %o0, %o3

                ! Divide by zero trap.  If it returns, return 0 (about as
                ! wrong as possible, but that is what SunOS does...).
                ta        0x2   !ST_DIV0
                retl
                clr        %o0

1:
        cmp        %o3, %o5                ! if %o1 exceeds %o0, done
        blu        got_result                ! (and algorithm fails otherwise)
        clr        %o2
        sethi        %hi(1 << (32 - 4 - 1)), %g1
        cmp        %o3, %g1
        blu        not_really_big
        clr        %o4

        ! Here the dividend is >= 2**(31-N) or so.  We must be careful here,
        ! as our usual N-at-a-shot divide step will cause overflow and havoc.
        ! The number of bits in the result here is N*ITER+SC, where SC <= N.
        ! Compute ITER in an unorthodox manner: know we need to shift V into
        ! the top decade: so do not even bother to compare to R.
        1:
                cmp        %o5, %g1
                bgeu        3f
                mov        1, %g2
                sll        %o5, 4, %o5
                b        1b
                add        %o4, 1, %o4

        ! Now compute %g2.
        2:        addcc        %o5, %o5, %o5
                bcc        not_too_big
                add        %g2, 1, %g2

                ! We get here if the %o1 overflowed while shifting.
                ! This means that %o3 has the high-order bit set.
                ! Restore %o5 and subtract from %o3.
                sll        %g1, 4, %g1        ! high order bit
                srl        %o5, 1, %o5                ! rest of %o5
                add        %o5, %g1, %o5
                b        do_single_div
                sub        %g2, 1, %g2

        not_too_big:
        3:        cmp        %o5, %o3
                blu        2b
                nop
                be        do_single_div
                nop
        /* NB: these are commented out in the V8-SPARC manual as well */
        /* (I do not understand this) */
        ! %o5 > %o3: went too far: back up 1 step
        !        srl        %o5, 1, %o5
        !        dec        %g2
        ! do single-bit divide steps
        !
        ! We have to be careful here.  We know that %o3 >= %o5, so we can do the
        ! first divide step without thinking.  BUT, the others are conditional,
        ! and are only done if %o3 >= 0.  Because both %o3 and %o5 may have the high-
        ! order bit set in the first step, just falling into the regular
        ! division loop will mess up the first time around.
        ! So we unroll slightly...
        do_single_div:
                subcc        %g2, 1, %g2
                bl        end_regular_divide
                nop
                sub        %o3, %o5, %o3
                mov        1, %o2
                b        end_single_divloop
                nop
        single_divloop:
                sll        %o2, 1, %o2
                bl        1f
                srl        %o5, 1, %o5
                ! %o3 >= 0
                sub        %o3, %o5, %o3
                b        2f
                add        %o2, 1, %o2
        1:        ! %o3 < 0
                add        %o3, %o5, %o3
                sub        %o2, 1, %o2
        2:
        end_single_divloop:
                subcc        %g2, 1, %g2
                bge        single_divloop
                tst        %o3
                b,a        end_regular_divide

not_really_big:
1:
        sll        %o5, 4, %o5
        cmp        %o5, %o3
        bleu        1b
        addcc        %o4, 1, %o4
        be        got_result
        sub        %o4, 1, %o4

        tst        %o3        ! set up for initial iteration
divloop:
        sll        %o2, 4, %o2
                ! depth 1, accumulated bits 0
        bl        L1.16
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 2, accumulated bits 1
        bl        L2.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 3, accumulated bits 3
        bl        L3.19
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 7
        bl        L4.23
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (7*2+1), %o2
L4.23:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (7*2-1), %o2
        
L3.19:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 5
        bl        L4.21
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (5*2+1), %o2
        
L4.21:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (5*2-1), %o2
        
L2.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 3, accumulated bits 1
        bl        L3.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 3
        bl        L4.19
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (3*2+1), %o2
        
L4.19:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (3*2-1), %o2
        
L3.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits 1
        bl        L4.17
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (1*2+1), %o2
        
L4.17:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (1*2-1), %o2
        
L1.16:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 2, accumulated bits -1
        bl        L2.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 3, accumulated bits -1
        bl        L3.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -1
        bl        L4.15
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-1*2+1), %o2
        
L4.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-1*2-1), %o2
        
L3.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -3
        bl        L4.13
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-3*2+1), %o2
        
L4.13:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-3*2-1), %o2
        
L2.15:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 3, accumulated bits -3
        bl        L3.13
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -5
        bl        L4.11
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-5*2+1), %o2
        
L4.11:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-5*2-1), %o2
        
L3.13:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        ! depth 4, accumulated bits -7
        bl        L4.9
        srl        %o5,1,%o5
        ! remainder is positive
        subcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-7*2+1), %o2
        
L4.9:
        ! remainder is negative
        addcc        %o3,%o5,%o3
        b        9f
        add        %o2, (-7*2-1), %o2
        
        9:
end_regular_divide:
        subcc        %o4, 1, %o4
        bge        divloop
        tst        %o3
        bl,a        got_result
        ! non-restoring fixup here (one instruction only!)
        add        %o3, %o1, %o3

got_result:
        ! check to see if answer should be < 0
        tst        %g3
        bl,a        1f
        sub %g0, %o3, %o3
1:
        retl
        mov %o3, %o0

#endif