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;$Header: /home/dashley/cvsrep/e3ft_gpl01/e3ft_gpl01/dtaipubs/cron/2007/cpu08div16by16a/div16e16x16.s,v 1.4 2007/05/26 17:48:46 dashley Exp $:
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;------------------------------------------------------------------------------
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xdef c_udiv, c_umod
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xref.b c_reg
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.dcall "9,0,c_umod"
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.dcall "7,0,c_udiv"
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;
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;Function Entry Conditions:
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; NH in C_REG (a zero-page, i.e. direct addressing variable)
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; NL in A
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; DH in H
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; DL in X
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;
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;Function Exit Conditions
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; QH in C_REG
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; QL in A
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; RH in H
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; RL in X
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; Stack pointer (naturally) restored.
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; No restrictions on condition codes (can be arbitrarily changed).
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; No restrictions on other CPU registers (can be arbitrarily changed).
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;
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c_udiv:
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psha ; save NL
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pshh ; DH on the stack
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tst 1,sp ; test zero
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bne full ; full division
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;
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;Short divisor division.
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;Divisor <= 255 (and may also be zero).
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;
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;Note to MJB: your staff did a better job on this than
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;me. I had 48 clocks, this path seems to be 45.
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;
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pulh ; clean stack
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lda c_reg ; NH in A
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clrh ; 1st divide 8 X 8
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div ; divide (H:A)/(X) -> A with remainder H
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sta c_reg ; QH in place
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pula ; complete dividend with NL
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div ; divide
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pshh ; move RL
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pulx ; in X
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clrh ; RH is zero
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rts ; and return
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;
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;Long Divisor Division
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;
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;Mathematical Assertions
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; Divisor >= 256 (and is definitely not zero).
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; Numerator can be any value.
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; Quotient <= 255.
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; Remainder may be two bytes.
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;
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;Entry Conditions
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; Stack frame already grown by 2 (and could use 1,SP and 2,SP
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; for intermediate results).
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; NH in C_REG (a zero-page, i.e. direct addressing variable)
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; NL in A and 2,SP
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; DH in H and 1,SP
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; DL in X
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;
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full:
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pshx ; save DL
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psha ; start with
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lda c_reg ; dividend
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psha
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;
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;Stack condition
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; 1,SP NH
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; 2,SP NL
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; 3,SP DL
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; 4,SP DH
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; 5,SP NL
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;Due to redundancy, can probably re-use 5,SP for
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;something else.
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;
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lda 4,sp ; DH
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;
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;Shift the numerator and denominator right until the denominator
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;just fits in one byte. At least one shift will be required,
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;because the denominator is known to be >= 256, hence the
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;do-while() rather than while() loop.
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;
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sbcl:
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lsr 1,sp
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ror 2,sp
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lsra
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rorx
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tsta
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bne sbcl
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;
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;DTA note: the loop above looks excellent. No real fat. The only
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;optimizations that seems possible are (a) to use c_reg to shift one byte
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;of the numerator in so that a 4-cycle rather than a 5-cycle instruction
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;is used, or (b) to use X rather than SP so that have 2 4-cycle rather than
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;2 5-cycle instructions.
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;
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;Do the initial digit estimation division. Knuth calls out the possibility of
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;an overflow, and that an overflow should be replaced by a quotient of 255.
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;Not sure if an overflow is possible. 0 <= N <= 65535, and 256 <= D <= 65535.
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;Assuming max N and min D after one shift gives 32767/128 = 255, still safe.
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;Need to be more rigorous about the math to be sure that overflow not possible
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;on the division. Note also that Knuth's algorithm calls for left-shifting to
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;normalize and for the creation of an extra digit (in our case, extra byte).
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;Because we have bounds on the divisor going in, right-shifting gives the same
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;effect for us. There may be something about the setup that changes Knuth's
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;assumptions that overflow is possible. Will need to revisit this and prove
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;it rigorously.
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;
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txa
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pulh
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pulx
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div
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;
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;Let:
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; N be the numerator in to the function.
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; D be the denominator in to the function.
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; Q be the actual quotient.
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; R be the actual remainder.
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; Z(0) be the first trial quotient we calculated.
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; Z(-1) be the trial quotient reduced by one.
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; Z(-2) be the trial quotient reduced by two.
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; Z() be the trial quotient we are working with at some
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; point in time.
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;
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;The original MJB approach was to reconstruct Z * D, then
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;subtract D up to twice to get Z(-1) * D and Z(-2) * D. That seemed to
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;bloat the code. Let's try instead (to save FLASH) to loop using
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;multiplication until N - (Z() * D) < D, or equivalently
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;that Z() * D > N - D. We've gotta use the first expression, however,
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;because the second would require either an exception case or 2's
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;complement arithmetic and a third byte if N < D.
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;
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;
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; +-------------------------------------+
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; | Entry Condition: have Z(0) |
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; | calculated. |
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; +-------------------------------------+
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; |
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; |
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; \|/
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; +-------------------------------------+/
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; | Calculate Z() * D |------+
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; +-------------------------------------+\ |
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; | |
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; | |
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; \|/ |
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; Y +-------------------------------------+ |
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; +-----| Z() * D > 65535 ? | |
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; | +-------------------------------------+ |
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; | | N |
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; | | |
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; | \|/ |
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; | +-------------------------------------+ |
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; | | Calculate N - Z() * D | |
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; | +-------------------------------------+ |
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; | | |
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; | | |
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; | \|/ |
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; | +-------------------------------------+ |
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; | | N - Z() * D < D ? |------------+
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; | +-------------------------------------+ | |
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; | | | |
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; | | | |
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; | \|/ | |
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; | \+-------------------------------------+ | |
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; +-----| Z() -- |------+ |
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; /+-------------------------------------+ |
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; |
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; +--------------------------------+
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; |
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; \|/
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; Done
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;
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;Stack Frame Setup
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; Keeping track of the stack frame is a lot of work before everything is
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; refined, so will just use the instructions and pretend the stack frame
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; has been enlarged via AIS. (Note that I've swapped DH and DL below--they
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; seemed to be out of order from before.)
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;
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; Imaginary Stack Setup
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; 1,SP NH
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; 2,SP NL
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; 3,SP DH
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; 4,SP DL
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; 5,SP Z(0)
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; 6,SP Z() * D, High Byte
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; 7,SP Z() * D, Low Byte
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; 8,SP N - Z() * D, High Byte
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; 9,SP N - Z() * D, Low Byte
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;
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;
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sta 5,sp ;Save Z(0).
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;
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nextz:
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;-------------------
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;Calculate Z() * D
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;-------------------
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ldx 4,sp
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mul ;Lower product in X:A.
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sta 7,sp
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stx 8,sp
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lda 5,sp
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ldx 3,sp
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mul ;Upper product in X:A.
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tstx
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bne p2big ;If the MSB of the upper product (byte 3) isn't zero,
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;this automatically implies that Z() is too big. It may
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;be possible to shave out this test ... requires more
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;analysis.
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add 6,sp
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bcs p2big ;If there is a carry out of adding the MSB in, this is
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;also an automatic cue that Z() is too big. It may
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;be possible to shave out this test ... requires more
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;analysis.
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sta 6,sp ;Put result back. (6,sp):(7,sp) now contains product,
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;and it definitely fits in 16 bits.
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;
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;-----------------------
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;Calculate N - Z() * D
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;-----------------------
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lda 2,sp
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sub 7,sp
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sta 9,sp
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lda 1,sp
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sub 6,sp
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sta 8,sp
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bcs p2big ;If N-Z()*D is negative, Z() is too big.
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;-------------------------------
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;Compare N - Z() * D against D
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;-------------------------------
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;Branches below are imaginary -- too lazy to look up the
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;instructions right now.
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;-------------------------------
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lda 3,sp
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cmp 8,sp
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bcc zwrong ;Need to look up right branches.
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lda 4,sp
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cmp 9,sp
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bcc zwrong ;Need to look up right branches.
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bra zright
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;-------------------------------
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;Z() --
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;-------------------------------
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p2big:
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zwrong:
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dec 5,sp
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lda 5,sp
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bra nextz
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;At this point, the remainder is definitely is (8,sp):(9,sp), and
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;the quotient is in 5,sp. Set up and return.
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zright:
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lda 8,sp
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psha
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pulh ;MSB of remainder in H.
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ldx 9,sp ;LSB of remainder in X.
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clr c_reg ;MSB of quotient zero.
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lda 5,sp ;LSB of quotient in A.
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ais #8 ;Adjust stack. Imaginary offset.
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rts ;And return
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;
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end
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;------------------------------------------------------------------------------
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;$Log: div16e16x16.s,v $
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;Revision 1.4 2007/05/26 17:48:46 dashley
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;Preliminary alternate solution in place.
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;
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;Revision 1.3 2007/05/26 16:13:10 dashley
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;Safety checkin after substantial edits.
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;
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;Revision 1.2 2007/05/25 23:34:31 dashley
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;Proposed Knuth solution integrated into version-control system.
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