785 lines
16 KiB
NASM
785 lines
16 KiB
NASM
/* This is an assembly language implementation of mulsi3, divsi3, and modsi3
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for the sparc processor.
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These routines are derived from the Sparc Architecture Manual, version 8,
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slightly edited to match the desired calling convention, and also to
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optimize them for our purposes. */
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#ifdef L_mulsi3
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.text
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.align 4
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.global .umul
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.proc 4
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.umul:
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or %o0, %o1, %o4 ! logical or of multiplier and multiplicand
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mov %o0, %y ! multiplier to Y register
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andncc %o4, 0xfff, %o5 ! mask out lower 12 bits
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be mul_shortway ! can do it the short way
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andcc %g0, %g0, %o4 ! zero the partial product and clear NV cc
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!
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! long multiply
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!
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mulscc %o4, %o1, %o4 ! first iteration of 33
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4 ! 32nd iteration
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mulscc %o4, %g0, %o4 ! last iteration only shifts
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! the upper 32 bits of product are wrong, but we do not care
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retl
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rd %y, %o0
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!
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! short multiply
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!
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mul_shortway:
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mulscc %o4, %o1, %o4 ! first iteration of 13
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4
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mulscc %o4, %o1, %o4 ! 12th iteration
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mulscc %o4, %g0, %o4 ! last iteration only shifts
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rd %y, %o5
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sll %o4, 12, %o4 ! left shift partial product by 12 bits
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srl %o5, 20, %o5 ! right shift partial product by 20 bits
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retl
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or %o5, %o4, %o0 ! merge for true product
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#endif
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#ifdef L_divsi3
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/*
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* Division and remainder, from Appendix E of the Sparc Version 8
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* Architecture Manual, with fixes from Gordon Irlam.
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*/
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/*
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* Input: dividend and divisor in %o0 and %o1 respectively.
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*
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* m4 parameters:
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* .div name of function to generate
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* div div=div => %o0 / %o1; div=rem => %o0 % %o1
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* true true=true => signed; true=false => unsigned
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*
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* Algorithm parameters:
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* N how many bits per iteration we try to get (4)
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* WORDSIZE total number of bits (32)
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*
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* Derived constants:
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* TOPBITS number of bits in the top decade of a number
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*
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* Important variables:
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* Q the partial quotient under development (initially 0)
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* R the remainder so far, initially the dividend
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* ITER number of main division loop iterations required;
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* equal to ceil(log2(quotient) / N). Note that this
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* is the log base (2^N) of the quotient.
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* V the current comparand, initially divisor*2^(ITER*N-1)
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*
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* Cost:
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* Current estimate for non-large dividend is
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* ceil(log2(quotient) / N) * (10 + 7N/2) + C
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* A large dividend is one greater than 2^(31-TOPBITS) and takes a
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* different path, as the upper bits of the quotient must be developed
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* one bit at a time.
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*/
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.global .udiv
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.align 4
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.proc 4
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.text
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.udiv:
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b ready_to_divide
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mov 0, %g3 ! result is always positive
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.global .div
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.align 4
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.proc 4
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.text
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.div:
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! compute sign of result; if neither is negative, no problem
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orcc %o1, %o0, %g0 ! either negative?
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bge ready_to_divide ! no, go do the divide
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xor %o1, %o0, %g3 ! compute sign in any case
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tst %o1
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bge 1f
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tst %o0
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! %o1 is definitely negative; %o0 might also be negative
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bge ready_to_divide ! if %o0 not negative...
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sub %g0, %o1, %o1 ! in any case, make %o1 nonneg
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1: ! %o0 is negative, %o1 is nonnegative
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sub %g0, %o0, %o0 ! make %o0 nonnegative
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ready_to_divide:
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! Ready to divide. Compute size of quotient; scale comparand.
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orcc %o1, %g0, %o5
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bne 1f
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mov %o0, %o3
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! Divide by zero trap. If it returns, return 0 (about as
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! wrong as possible, but that is what SunOS does...).
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ta 0x2 ! ST_DIV0
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retl
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clr %o0
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1:
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cmp %o3, %o5 ! if %o1 exceeds %o0, done
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blu got_result ! (and algorithm fails otherwise)
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clr %o2
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sethi %hi(1 << (32 - 4 - 1)), %g1
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cmp %o3, %g1
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blu not_really_big
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clr %o4
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! Here the dividend is >= 2**(31-N) or so. We must be careful here,
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! as our usual N-at-a-shot divide step will cause overflow and havoc.
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! The number of bits in the result here is N*ITER+SC, where SC <= N.
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! Compute ITER in an unorthodox manner: know we need to shift V into
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! the top decade: so do not even bother to compare to R.
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1:
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cmp %o5, %g1
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bgeu 3f
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mov 1, %g2
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sll %o5, 4, %o5
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b 1b
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add %o4, 1, %o4
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! Now compute %g2.
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2: addcc %o5, %o5, %o5
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bcc not_too_big
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add %g2, 1, %g2
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! We get here if the %o1 overflowed while shifting.
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! This means that %o3 has the high-order bit set.
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! Restore %o5 and subtract from %o3.
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sll %g1, 4, %g1 ! high order bit
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srl %o5, 1, %o5 ! rest of %o5
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add %o5, %g1, %o5
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b do_single_div
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sub %g2, 1, %g2
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not_too_big:
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3: cmp %o5, %o3
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blu 2b
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nop
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be do_single_div
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nop
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/* NB: these are commented out in the V8-Sparc manual as well */
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/* (I do not understand this) */
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! %o5 > %o3: went too far: back up 1 step
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! srl %o5, 1, %o5
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! dec %g2
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! do single-bit divide steps
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!
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! We have to be careful here. We know that %o3 >= %o5, so we can do the
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! first divide step without thinking. BUT, the others are conditional,
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! and are only done if %o3 >= 0. Because both %o3 and %o5 may have the high-
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! order bit set in the first step, just falling into the regular
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! division loop will mess up the first time around.
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! So we unroll slightly...
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do_single_div:
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subcc %g2, 1, %g2
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bl end_regular_divide
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nop
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sub %o3, %o5, %o3
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mov 1, %o2
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b end_single_divloop
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nop
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single_divloop:
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sll %o2, 1, %o2
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bl 1f
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srl %o5, 1, %o5
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! %o3 >= 0
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sub %o3, %o5, %o3
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b 2f
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add %o2, 1, %o2
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1: ! %o3 < 0
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add %o3, %o5, %o3
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sub %o2, 1, %o2
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2:
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end_single_divloop:
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subcc %g2, 1, %g2
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bge single_divloop
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tst %o3
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b,a end_regular_divide
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not_really_big:
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1:
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sll %o5, 4, %o5
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cmp %o5, %o3
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bleu 1b
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addcc %o4, 1, %o4
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be got_result
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sub %o4, 1, %o4
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tst %o3 ! set up for initial iteration
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divloop:
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sll %o2, 4, %o2
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! depth 1, accumulated bits 0
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bl L1.16
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 2, accumulated bits 1
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bl L2.17
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 3, accumulated bits 3
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bl L3.19
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 4, accumulated bits 7
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bl L4.23
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (7*2+1), %o2
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L4.23:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (7*2-1), %o2
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L3.19:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 4, accumulated bits 5
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bl L4.21
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (5*2+1), %o2
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L4.21:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (5*2-1), %o2
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L2.17:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 3, accumulated bits 1
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bl L3.17
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 4, accumulated bits 3
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bl L4.19
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (3*2+1), %o2
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L4.19:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (3*2-1), %o2
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L3.17:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 4, accumulated bits 1
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bl L4.17
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (1*2+1), %o2
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L4.17:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (1*2-1), %o2
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L1.16:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 2, accumulated bits -1
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bl L2.15
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 3, accumulated bits -1
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bl L3.15
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 4, accumulated bits -1
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bl L4.15
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (-1*2+1), %o2
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L4.15:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (-1*2-1), %o2
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L3.15:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 4, accumulated bits -3
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bl L4.13
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (-3*2+1), %o2
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L4.13:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (-3*2-1), %o2
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L2.15:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 3, accumulated bits -3
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bl L3.13
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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! depth 4, accumulated bits -5
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bl L4.11
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (-5*2+1), %o2
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L4.11:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (-5*2-1), %o2
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L3.13:
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! remainder is negative
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addcc %o3,%o5,%o3
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! depth 4, accumulated bits -7
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bl L4.9
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srl %o5,1,%o5
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! remainder is positive
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subcc %o3,%o5,%o3
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b 9f
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add %o2, (-7*2+1), %o2
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L4.9:
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! remainder is negative
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addcc %o3,%o5,%o3
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b 9f
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add %o2, (-7*2-1), %o2
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9:
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end_regular_divide:
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subcc %o4, 1, %o4
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bge divloop
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tst %o3
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bl,a got_result
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! non-restoring fixup here (one instruction only!)
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sub %o2, 1, %o2
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got_result:
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! check to see if answer should be < 0
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tst %g3
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bl,a 1f
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sub %g0, %o2, %o2
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1:
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retl
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mov %o2, %o0
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#endif
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#ifdef L_modsi3
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/* This implementation was taken from glibc:
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*
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* Input: dividend and divisor in %o0 and %o1 respectively.
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*
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* Algorithm parameters:
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* N how many bits per iteration we try to get (4)
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* WORDSIZE total number of bits (32)
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*
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* Derived constants:
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* TOPBITS number of bits in the top decade of a number
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*
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* Important variables:
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* Q the partial quotient under development (initially 0)
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* R the remainder so far, initially the dividend
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* ITER number of main division loop iterations required;
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* equal to ceil(log2(quotient) / N). Note that this
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* is the log base (2^N) of the quotient.
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* V the current comparand, initially divisor*2^(ITER*N-1)
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*
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* Cost:
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* Current estimate for non-large dividend is
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* ceil(log2(quotient) / N) * (10 + 7N/2) + C
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* A large dividend is one greater than 2^(31-TOPBITS) and takes a
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* different path, as the upper bits of the quotient must be developed
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* one bit at a time.
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*/
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.text
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.align 4
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.global .urem
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.proc 4
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.urem:
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b divide
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mov 0, %g3 ! result always positive
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.align 4
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.global .rem
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.proc 4
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.rem:
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! compute sign of result; if neither is negative, no problem
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orcc %o1, %o0, %g0 ! either negative?
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bge 2f ! no, go do the divide
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mov %o0, %g3 ! sign of remainder matches %o0
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tst %o1
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bge 1f
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tst %o0
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! %o1 is definitely negative; %o0 might also be negative
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bge 2f ! if %o0 not negative...
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sub %g0, %o1, %o1 ! in any case, make %o1 nonneg
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1: ! %o0 is negative, %o1 is nonnegative
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sub %g0, %o0, %o0 ! make %o0 nonnegative
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2:
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! Ready to divide. Compute size of quotient; scale comparand.
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divide:
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orcc %o1, %g0, %o5
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bne 1f
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mov %o0, %o3
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! Divide by zero trap. If it returns, return 0 (about as
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! wrong as possible, but that is what SunOS does...).
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ta 0x2 !ST_DIV0
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retl
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clr %o0
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1:
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cmp %o3, %o5 ! if %o1 exceeds %o0, done
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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
|
|
|