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   /* This file is generated from divrem.m4; DO NOT EDIT! */
/*
 * 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:
 *  .rem        name of function to generate
 *  rem         rem=div => %o0 / %o1; rem=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.
 */



#include <sys/syscall.h>


.global   .rem;
.align 4;
.type  .rem ,@function; 

.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.
        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      0x02
                retl
                clr     %o0

1:
        cmp     %o3, %o5                        ! if %o1 exceeds %o0, done
        blu     .Lgot_result            ! (and algorithm fails otherwise)
        clr     %o2
        sethi   %hi(1 << (32 - 4 - 1)), %g1
        cmp     %o3, %g1
        blu     .Lnot_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     .Lnot_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       .Ldo_single_div
                sub     %g2, 1, %g2

        .Lnot_too_big:
        3:      cmp     %o5, %o3
                blu     2b
                nop
                be      .Ldo_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...
        .Ldo_single_div:
                subcc   %g2, 1, %g2
                bl      .Lend_regular_divide
                nop
                sub     %o3, %o5, %o3
                mov     1, %o2
                b       .Lend_single_divloop
                nop
        .Lsingle_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:
        .Lend_single_divloop:
                subcc   %g2, 1, %g2
                bge     .Lsingle_divloop
                tst     %o3
                b,a     .Lend_regular_divide

.Lnot_really_big:
1:
        sll     %o5, 4, %o5
        cmp     %o5, %o3
        bleu    1b
        addcc   %o4, 1, %o4
        be      .Lgot_result
        sub     %o4, 1, %o4

        tst     %o3     ! set up for initial iteration
.Ldivloop:
        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:
.Lend_regular_divide:
        subcc   %o4, 1, %o4
        bge     .Ldivloop
        tst     %o3
        bl,a    .Lgot_result
        ! non-restoring fixup here (one instruction only!)
        add     %o3, %o1, %o3


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

.size .rem,.-.rem;