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%$Header: /home/dashley/cvsrep/e3ft_gpl01/e3ft_gpl01/dtaipubs/cron/2007/cpu08div16by16a/cpu08div16by16a.tex,v 1.12 2007/05/26 18:25:46 dashley Exp $
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%
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\documentclass[letterpaper,10pt,titlepage]{article}
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%\pagestyle{headings}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage[ansinew]{inputenc}
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\usepackage[OT1]{fontenc}
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\usepackage{graphicx}
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%\usepackage{makeidx}
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%
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%Contributed by Robin Fairbairns of the comp.text.tex newsgroup. I have no
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%idea how it works ... but it gives a working \bdiv.
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%
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\makeatletter
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\def\bdiv{%
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\nonscript\mskip-\medmuskip\mkern5mu%
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\mathbin{\operator@font div}\penalty900\mkern5mu%
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\nonscript\mskip-\medmuskip}
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\makeatother
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%
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%
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\begin{document}
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\title{Notes on Economical 16/16 = 16 Unsigned Integer Division on the Freescale CPU08}
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\author{\vspace{1cm}\\David T. Ashley\\\texttt{dta@e3ft.com}\\\vspace{1cm}}
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\date{\vspace*{8mm}\small{Version Control $ $Revision: 1.12 $ $ \\
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Version Control $ $Date: 2007/05/26 18:25:46 $ $ (UTC) \\
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$ $RCSfile: cpu08div16by16a.tex,v $ $ \\
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\LaTeX{} Compilation Date: \today{}}}
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\maketitle
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%
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\begin{abstract}
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This document contains my notes on attempting to optimize
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the 16 = 16/16 unsigned integer division subroutine distributed with
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a Freescale CPU08 compiler. I attempted to replace the standard
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shift-compare-subtract algorithm with Knuth's algorithm.
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In the end, the attempt was mostly a failure. There seemed to be no way
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to gain the execution time advantages of Knuth's algorithm without increasing
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FLASH size. (The goal would have been to obtain more favorable
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execution time \emph{and} FLASH size, rather than a tradeoff.)
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\end{abstract}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Introduction and Overview}
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\label{siov0}
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The fundamental goal of practical computer integer arithmetic is to obtain
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an exact result using machine instructions that are as fast and compact as
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possible.
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For three of the four fundamental operations (addition, subtraction, and
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multiplication), it is intuitively obvious to most programmers how to
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use existing machine instructions to operate on operands that are larger
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than the machine instructions can accommodate.
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The fourth fundamental operation (division), however, is less well understood
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by a typical programmer than the other three. It is not obvious to most
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programmers how to use machine division instructions to divide integers larger
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than the native machine division instructions will accommodate.
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In 2007, I noticed that the integer division library functions associated with
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a particular \emph{C} compiler were of the shift-compare-subtract variety.
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As the CPU08 has a 16/16=8 native division instruction, I suspected but was not
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sure that the 16/16=16 division function could be improved.
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Note that for a library function used by a \emph{C} compiler, it may not
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be necessary (although it may be desirable) to calculate the quotient and the
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remainder at the same time. An algorithm that calculates the quotient but
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does not calculate the remainder, or vice-versa, may be acceptable for
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use by compiled code.
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Although I haven't examined the \emph{C} standards, I doubt that there is
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any special requirement for the quotient or remainder calculated when
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the denominator is zero. The only requirement is probably that the calculation
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not continue indefinitely (i.e. not an infinite loop).
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Nomenclature and Notation}
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\label{snom0}
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I use the nomenclature ``16/16=16 division'' to denote a division function that
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accepts a 16-bit unsigned numerator and a 16-bit unsigned denominator; and
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produces either/both a 16-bit quotient and a 16-bit remainder.
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I use $n$ to denote the numerator, $d$ to denote the denominator, $q$ to denote
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the integer quotient, and $r$ to denote the remainder.
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Any of the 16-bit quantities can be subscripted with ``H'' or ``L'' to denote
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the most-significant or least-significant byte. For example,
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\begin{equation}
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\label{eq:snom0:01}
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n = 2^8 n_H + n_L .
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\end{equation}
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Within any of the 16-bit quantities, bits are subscripted in the traditional fashion.
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For example,
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\begin{equation}
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\label{eq:snom0:02}
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n = \sum_{i=0}^{15} 2^i n_i .
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\end{equation}
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Within any of the 8-bit quantities, bits are also subscripted in the traditional fashion.
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For example,
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\begin{equation}
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\label{eq:snom0:03}
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n = 2^8 \sum_{i=0}^{7} 2^i n_{H_i} + \sum_{i=0}^{7} 2^i n_{L_i}
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= \sum_{i=0}^{7} 2^{i+8} n_{H_i} + \sum_{i=0}^{7} 2^i n_{L_i} .
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\end{equation}
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Because only non-negative integers are involved, \emph{floor()} and
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\emph{div} are used interchangeably, i.e.
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\begin{equation}
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\label{eq:snom0:04}
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\left\lfloor \frac{a}{b} \right\rfloor = a \bdiv b, \;\; a,b \geq 0 .
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\end{equation}
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An identity that is frequently used in this document is
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\begin{eqnarray}
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\label{eq:snom0:05}
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\frac{a}{b} & = & a \bdiv b + \frac{a \bmod b}{b} \\
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\label{eq:snom0:06}
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& = & \left\lfloor \frac{a}{b} \right\rfloor + \frac{a \bmod b}{b} .
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\end{eqnarray}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Analysis of Existing Code}
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\label{saec0}
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This section presents an analysis of the behavior and characteristics of
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the existing code.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Listing}
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\label{saec0:slec0}
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The existing code is reproduced below, with line numbers.
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\begin{verbatim}
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001: c_umod:
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002: bsr c_udiv ; divide
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003: pshh ; transfer MSB
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004: pula ; in place
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005: sta c_reg
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006: txa ; LSB in place
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007: rts ; and return
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008: ;
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009: c_udiv:
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010: psha ; save NL
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011: pshh ; DH on the stack
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012: tst 1,sp ; test zero
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013: bne full ; full division
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014: pulh ; clean stack
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015: cpx c_reg ; compare DL with NH
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016: bls half ; half division
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017: lda c_reg ; NH
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018: psha ; in
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019: pulh ; H
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020: pula ; NL in A
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021: div ; DL in X, divide
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022: clr c_reg ; QH is zero
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023: fini:
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024: pshh ; move RL
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025: pulx ; in X
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026: clrh ; RH is zero
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027: rts ; and return
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028: half:
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029: lda c_reg ; NH in A
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030: clrh ; 1st divide 8 X 8
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031: div ; divide
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032: sta c_reg ; QH in place
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033: pula ; complete dividend with NL
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034: div ; divide
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035: bra fini ; complete remainder
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036: full:
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037: pshx ; save DL
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038: ldx #8 ; counter
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039: clra ; Extention
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040: pshx ; save on
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041: psha ; the stack
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042: tsx ; stack addressed by H:X
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043: bcl:
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044: lsl 4,x ; shift E:NH:NL
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045: rol c_reg ; left
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046: rola
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047: sta 0,x ; store E
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048: cmp 3,x ; compare DH
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049: blo next ; too low, continue
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050: bne ok ; ok to subtract
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051: lda c_reg ; compare NH
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052: sub 2,x ; with DL
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053: bhs ok2 ; ok, complete result
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054: lda 0,x ; restore value
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055: bra next ; and continue
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056: ok:
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057: lda c_reg ; substract D
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058: sub 2,x ; from E:NH
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059: ok2:
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060: sta c_reg ; in place
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061: lda 0,x
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062: sbc 3,x
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063: inc 4,x ; set result bit
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064: next:
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065: dbnz 1,x,bcl ; count down and loop
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066: sta 0,x ; store E
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067: pulh ; RH in place
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068: ais #3 ; clean up stack
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069: ldx c_reg ; RL in place
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070: pula ; QL in place
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071: clr c_reg ; QH is zero
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072: rts ; and return
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\end{verbatim}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Algorithmic Analysis}
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\label{saec0:saan0}
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The algorithm is divided into two cases:
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\begin{itemize}
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\item Case I: $d_H = 0$ (lines 14-35).
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\item Case II: $d_H > 0$ (lines 36-72).
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\end{itemize}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection{Case I}
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\label{saec0:saan0:scas0}
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In the case of $d_H = 0$, the division is 16/8:
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\begin{eqnarray}
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\label{eq:saec0:saan0:scas0:01}
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\frac{2^8 n_H + n_L}{d_L} & = & \frac{2^8 n_H}{d_L} + \frac{n_L}{d_L} \\
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\label{eq:saec0:saan0:scas0:02}
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& = & 2^8 \left( n_H \bdiv d_L + \frac{n_H \bmod d_L}{d_L} \right) + \frac{n_L}{d_L} \\
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\label{eq:saec0:saan0:scas0:03}
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& = & 2^8 (n_H \bdiv d_L) + \frac{2^8 (n_H \bmod d_L) + n_L}{d_L}
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\end{eqnarray}
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(\ref{eq:saec0:saan0:scas0:03}) is an exact expression involving rational numbers.
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However, we don't want to calculate the left side of (\ref{eq:saec0:saan0:scas0:03});
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rather, we wish to calculate its \emph{floor()}\@. Applying the
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\emph{floor()} function to both sides of (\ref{eq:saec0:saan0:scas0:03}) yields:
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\begin{equation}
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\label{eq:saec0:saan0:scas0:04}
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\left\lfloor \frac{2^8 n_H + n_L}{d_L} \right\rfloor
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=
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2^8 (n_H \bdiv d_L) + \left\lfloor \frac{2^8 (n_H \bmod d_L) + n_L}{d_L} \right\rfloor .
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\end{equation}
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Note that (\ref{eq:saec0:saan0:scas0:04}) is in a form
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that can be readily evaluated using a processor with 16/8 division
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capability; so long as
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\begin{equation}
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\label{eq:saec0:saan0:scas0:05}
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\frac{2^8 (n_H \bmod d_L) + n_L}{d_L} < 2^8 ,
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\end{equation}
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\noindent{}a fact that can be easily verified by the reader.
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(\ref{eq:saec0:saan0:scas0:04}) can be readily evaluated by a processor with
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16/8 division capability because such a division instruction always provides
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both quotient and remainder. It is easy to see that (\ref{eq:saec0:saan0:scas0:04})
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can be evaluated with a division, a re-staging of bytes, and a second division.
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If (\ref{eq:saec0:saan0:scas0:04}) is evaluated as suggested, it needs to be verified
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whether the remainder of the second division is the same as the remainder
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of the larger division, i.e.
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\begin{equation}
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\label{eq:saec0:saan0:scas0:06}
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(2^8 n_H + n_L) \bmod d_L =? ((2^8 \bmod d_L) + n_L) \bmod d_L.
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\end{equation}
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The question of whether (\ref{eq:saec0:saan0:scas0:06}) is an
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equality is the question of whether
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\begin{equation}
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\label{eq:saec0:saan0:scas0:07}
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ka \bmod b = (k (a \bmod b)) \bmod b .
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\end{equation}
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In order to prove or disprove (\ref{eq:saec0:saan0:scas0:07}),
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decompose $a$ into $ib + j$. Then, since $kib \bmod b = 0$,
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\begin{eqnarray}
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\label{eq:saec0:saan0:scas0:08}
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k (ib + j) \bmod b & = & k j \bmod b \\
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\label{eq:saec0:saan0:scas0:09}
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k j \bmod b & = & k j \bmod b .
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\end{eqnarray}
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Thus, if (\ref{eq:saec0:saan0:scas0:04}) is evaluated as suggested (with
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two divisions), the final remainder will be the same as the remainder for the
|
| 317 |
|
|
original division. (\ref{eq:saec0:saan0:scas0:04}) will, in fact, deliver both
|
| 318 |
|
|
the quotient and remainder economically.
|
| 319 |
|
|
|
| 320 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 321 |
|
|
|
| 322 |
|
|
\subsubsection{Case II}
|
| 323 |
|
|
\label{saec0:saan0:scas1}
|
| 324 |
|
|
|
| 325 |
|
|
The case of $d_H > 0$ (\S{}\ref{saec0:slec0}, lines 36-72) is conventional
|
| 326 |
|
|
shift-compare-subtract division. Only eight iterations of the loop are required
|
| 327 |
|
|
because with $d_H > 0$, $d \geq 2^8$, and $n/d < 2^8$.
|
| 328 |
|
|
|
| 329 |
|
|
|
| 330 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 331 |
|
|
|
| 332 |
|
|
\subsection{Timing Analysis}
|
| 333 |
|
|
\label{saec0:stan0}
|
| 334 |
|
|
|
| 335 |
|
|
The code of \S{}\ref{saec0:slec0} is reproduced below, with instruction
|
| 336 |
|
|
timing (number of clocks) and FLASH requirements (number of bytes) added
|
| 337 |
|
|
as (clocks/bytes). It was determined that \emph{c\_reg} resides
|
| 338 |
|
|
in zero-page (i.e. \emph{direct}) memory.
|
| 339 |
|
|
|
| 340 |
|
|
\begin{verbatim}
|
| 341 |
|
|
001: c_umod:
|
| 342 |
|
|
002: bsr c_udiv 4/2 ; divide
|
| 343 |
|
|
003: pshh ; transfer MSB
|
| 344 |
|
|
004: pula 2/1 ; in place
|
| 345 |
|
|
005: sta c_reg 3/2
|
| 346 |
|
|
006: txa 1/1 ; LSB in place
|
| 347 |
|
|
007: rts 4/1 ; and return
|
| 348 |
|
|
008: ;
|
| 349 |
|
|
009: c_udiv:
|
| 350 |
|
|
010: psha 2/1 ; save NL
|
| 351 |
|
|
011: pshh 2/1 ; DH on the stack
|
| 352 |
|
|
012: tst 1,sp 4/3 ; test zero
|
| 353 |
|
|
013: bne full 3/2 ; full division
|
| 354 |
|
|
014: pulh 2/1 ; clean stack
|
| 355 |
|
|
015: cpx c_reg 3/2 ; compare DL with NH
|
| 356 |
|
|
016: bls half 3/2 ; half division
|
| 357 |
|
|
017: lda c_reg 3/2 ; NH
|
| 358 |
|
|
018: psha 2/1 ; in
|
| 359 |
|
|
019: pulh 2/1 ; H
|
| 360 |
|
|
020: pula 2/1 ; NL in A
|
| 361 |
|
|
021: div 7/1 ; DL in X, divide
|
| 362 |
|
|
022: clr c_reg 3/2 ; QH is zero
|
| 363 |
|
|
023: fini:
|
| 364 |
|
|
024: pshh 2/1 ; move RL
|
| 365 |
|
|
025: pulx 2/1 ; in X
|
| 366 |
|
|
026: clrh 1/1 ; RH is zero
|
| 367 |
|
|
027: rts 4/1 ; and return
|
| 368 |
|
|
028: half:
|
| 369 |
|
|
029: lda c_reg 3/2 ; NH in A
|
| 370 |
|
|
030: clrh 1/1 ; 1st divide 8 X 8
|
| 371 |
|
|
031: div 7/1 ; divide
|
| 372 |
|
|
032: sta c_reg 3/2 ; QH in place
|
| 373 |
|
|
033: pula 2/1 ; complete dividend with NL
|
| 374 |
|
|
034: div 7/1 ; divide
|
| 375 |
|
|
035: bra fini 3/2 ; complete remainder
|
| 376 |
|
|
036: full:
|
| 377 |
|
|
037: pshx 2/1 ; save DL
|
| 378 |
|
|
038: ldx #8 2/2 ; counter
|
| 379 |
|
|
039: clra 1/1 ; Extention
|
| 380 |
|
|
040: pshx 2/1 ; save on
|
| 381 |
|
|
041: psha 2/1 ; the stack
|
| 382 |
|
|
042: tsx 2/1 ; stack addressed by H:X
|
| 383 |
|
|
043: bcl:
|
| 384 |
|
|
044: lsl 4,x 4/2 ; shift E:NH:NL
|
| 385 |
|
|
045: rol c_reg 4/2 ; left
|
| 386 |
|
|
046: rola 1/1
|
| 387 |
|
|
047: sta 0,x 2/1 ; store E
|
| 388 |
|
|
048: cmp 3,x 3/2 ; compare DH
|
| 389 |
|
|
049: blo next 3/2 ; too low, continue
|
| 390 |
|
|
050: bne ok 3/2 ; ok to subtract
|
| 391 |
|
|
051: lda c_reg 3/2 ; compare NH
|
| 392 |
|
|
052: sub 2,x 3/2 ; with DL
|
| 393 |
|
|
053: bhs ok2 3/2 ; ok, complete result
|
| 394 |
|
|
054: lda 0,x 2/1 ; restore value
|
| 395 |
|
|
055: bra next 3/2 ; and continue
|
| 396 |
|
|
056: ok:
|
| 397 |
|
|
057: lda c_reg 3/2 ; substract D
|
| 398 |
|
|
058: sub 2,x 3/2 ; from E:NH
|
| 399 |
|
|
059: ok2:
|
| 400 |
|
|
060: sta c_reg 3/2 ; in place
|
| 401 |
|
|
061: lda 0,x 2/1
|
| 402 |
|
|
062: sbc 3,x 3/2
|
| 403 |
|
|
063: inc 4,x 3/2 ; set result bit
|
| 404 |
|
|
064: next:
|
| 405 |
|
|
065: dbnz 1,x,bcl 5/3 ; count down and loop
|
| 406 |
|
|
066: sta 0,x 2/1 ; store E
|
| 407 |
|
|
067: pulh 2/1 ; RH in place
|
| 408 |
|
|
068: ais #3 2/2 ; clean up stack
|
| 409 |
|
|
069: ldx c_reg 3/2 ; RL in place
|
| 410 |
|
|
070: pula 2/1 ; QL in place
|
| 411 |
|
|
071: clr c_reg 3/2 ; QH is zero
|
| 412 |
|
|
072: rts 4/1 ; and return
|
| 413 |
|
|
\end{verbatim}
|
| 414 |
|
|
|
| 415 |
|
|
There are three distinct timing cases for the \emph{c\_udiv} function:
|
| 416 |
|
|
|
| 417 |
|
|
\begin{enumerate}
|
| 418 |
|
|
\item $d_H = 0$ and $n_H < d_L$:
|
| 419 |
|
|
47 clocks are required, representing straight flow of the
|
| 420 |
|
|
instructions from line 10 through line 27.
|
| 421 |
|
|
\item $d_H = 0$ and $n_H \geq d_L$: 54 clocks are required.
|
| 422 |
|
|
\item $d_H > 0$ and every bit of the quotient is 1, in which case
|
| 423 |
|
|
400 clocks are required. This represents 22 clocks up through
|
| 424 |
|
|
line 42, 45 clocks $\times$ 8 in the lines from 43 through 65, and
|
| 425 |
|
|
18 clocks in the lines from 66 through 72.
|
| 426 |
|
|
\end{enumerate}
|
| 427 |
|
|
|
| 428 |
|
|
|
| 429 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 430 |
|
|
|
| 431 |
|
|
\subsection{FLASH/RAM Consumption Analysis}
|
| 432 |
|
|
\label{saec0:sfra0}
|
| 433 |
|
|
|
| 434 |
|
|
From \S{}\ref{saec0:stan0}, 93 bytes of FLASH are used. Only one byte
|
| 435 |
|
|
of RAM is used (\emph{c\_reg}, probably shared with other functions
|
| 436 |
|
|
as well).
|
| 437 |
|
|
|
| 438 |
|
|
|
| 439 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 440 |
|
|
|
| 441 |
|
|
\section{Potential Optimizations}
|
| 442 |
|
|
\label{spop0}
|
| 443 |
|
|
|
| 444 |
|
|
|
| 445 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 446 |
|
|
|
| 447 |
|
|
\subsection{Potential Case I Optimizations}
|
| 448 |
|
|
\label{spop0:scas0}
|
| 449 |
|
|
|
| 450 |
|
|
This section corresponds to \emph{Case I} of \S{}\ref{saec0:saan0:scas0}.
|
| 451 |
|
|
|
| 452 |
|
|
The most obvious observation about the code (\S{}\ref{saec0:slec0}) is that
|
| 453 |
|
|
division instructions are very inexpensive on the CPU08---7 clock cycles,
|
| 454 |
|
|
or about 2 typical instructions. Branching based on
|
| 455 |
|
|
$n_H \geq d_L$
|
| 456 |
|
|
(\S{}\ref{saec0:slec0}, lines 15-16)
|
| 457 |
|
|
may cost more in the test, the branch, and in other
|
| 458 |
|
|
data transfer overhead than is saved. It may make sense to apply the
|
| 459 |
|
|
full formula in (\ref{eq:saec0:saan0:scas0:04}) in all cases where
|
| 460 |
|
|
$d_H = 0$.
|
| 461 |
|
|
|
| 462 |
|
|
When $d_H = 0$, and if one assumes normal distribution of data, the expected
|
| 463 |
|
|
value of execution time is about $(47+54)/2 = 50.5$ clocks.\footnote{If the
|
| 464 |
|
|
data is assumed normally distributed, $n_H$ has about a 0.5 probability of being at least as large
|
| 465 |
|
|
as $d_L$.}
|
| 466 |
|
|
|
| 467 |
|
|
The code below combines two of the three timing cases into one by ignoring the
|
| 468 |
|
|
relationship between $n_H$ and $d_L$.
|
| 469 |
|
|
|
| 470 |
|
|
\begin{verbatim}
|
| 471 |
|
|
;Condition at function entry:
|
| 472 |
|
|
;N_H in 1,SP
|
| 473 |
|
|
;N_L in A
|
| 474 |
|
|
;D_H in H
|
| 475 |
|
|
;D_L in X
|
| 476 |
|
|
;
|
| 477 |
|
|
;Condition at function exit:
|
| 478 |
|
|
;Q_H in c_reg
|
| 479 |
|
|
;Q_L in A
|
| 480 |
|
|
;R_H in H
|
| 481 |
|
|
;R_L in X
|
| 482 |
|
|
;
|
| 483 |
|
|
c_udiv:
|
| 484 |
|
|
psha 2/1 ; save NL
|
| 485 |
|
|
pshh 2/1 ; DH on the stack
|
| 486 |
|
|
tst 1,sp 4/3 ; test zero
|
| 487 |
|
|
bne full 3/2 ; full division
|
| 488 |
|
|
;
|
| 489 |
|
|
;From here on we're committed to the division with
|
| 490 |
|
|
;arbitary numerator, and denominator <= 255.
|
| 491 |
|
|
; N_H at 3,sp
|
| 492 |
|
|
; N_L at 2,sp
|
| 493 |
|
|
; D_H at 1,sp
|
| 494 |
|
|
;
|
| 495 |
|
|
clrh 1/1
|
| 496 |
|
|
lda 3,sp 4/3 ; H:A now contains N_H
|
| 497 |
|
|
div 7/1 ; divide
|
| 498 |
|
|
sta c_reg 3/2 ; QH in place
|
| 499 |
|
|
lda 2,sp 4/3 ; complete dividend with NL
|
| 500 |
|
|
div 7/1 ; divide. Q_L in A, R_L in H
|
| 501 |
|
|
pshh 2/1 ; move RL
|
| 502 |
|
|
pulx 2/1 ; in X
|
| 503 |
|
|
clrh 1/1 ; RH is zero
|
| 504 |
|
|
ais #3 2/2 ; clean stack
|
| 505 |
|
|
rts 4/1 ; and return
|
| 506 |
|
|
\end{verbatim}
|
| 507 |
|
|
|
| 508 |
|
|
Although the code does raise the minimum execution time from 47 to
|
| 509 |
|
|
48 clocks:
|
| 510 |
|
|
|
| 511 |
|
|
\begin{itemize}
|
| 512 |
|
|
\item It lowers the expected value of the $d_H=0$ execution time from
|
| 513 |
|
|
50.5 to 48 clocks.
|
| 514 |
|
|
\item It saves approximately 15 bytes of FLASH.
|
| 515 |
|
|
\end{itemize}
|
| 516 |
|
|
|
| 517 |
|
|
This optimization is recommended.
|
| 518 |
|
|
|
| 519 |
|
|
|
| 520 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 521 |
|
|
|
| 522 |
|
|
\subsection{Potential Case II Optimizations}
|
| 523 |
|
|
\label{spop0:scas1}
|
| 524 |
|
|
|
| 525 |
|
|
This section corresponds to \emph{Case II} of \S{}\ref{saec0:saan0:scas1}.
|
| 526 |
|
|
|
| 527 |
|
|
I sent an e-mail to an engineer at the compiler manufacturer indicating
|
| 528 |
|
|
that:
|
| 529 |
|
|
|
| 530 |
|
|
\begin{itemize}
|
| 531 |
|
|
\item I believed \emph{Case I} could be optimized as indicated earlier
|
| 532 |
|
|
in the document.
|
| 533 |
|
|
\item I believed \emph{Case II} could be optimized by applying Knuth's
|
| 534 |
|
|
algorithm.
|
| 535 |
|
|
\end{itemize}
|
| 536 |
|
|
|
| 537 |
|
|
The reply I received from the compiler manufacturer was that:
|
| 538 |
|
|
|
| 539 |
|
|
\begin{itemize}
|
| 540 |
|
|
\item There was agreeement about \emph{Case I}.
|
| 541 |
|
|
\item \emph{Case II} may be a little faster using Knuth's algorithm, but would
|
| 542 |
|
|
definitely be larger (code used to evaluate the application of Knuth's
|
| 543 |
|
|
algorithm was also provided).
|
| 544 |
|
|
\end{itemize}
|
| 545 |
|
|
|
| 546 |
|
|
In the test code provided by the compiler manufacturer, the approach used was
|
| 547 |
|
|
to obtain a trial quotient and then to subtract the divisor from the remainder
|
| 548 |
|
|
up to twice to adjust the quotient up to 2 counts downward (Knuth's algorithm).
|
| 549 |
|
|
|
| 550 |
|
|
I did try a different approach (to iterate on the quotient and to reconstruct
|
| 551 |
|
|
$q \times d$ with decreasing $q$). This approach promised to be slightly more
|
| 552 |
|
|
compact because $q \times d$ reconstruction was reutilized. However, it worked out
|
| 553 |
|
|
to occupy about 153 bytes rather than 103 for the shift-compare-subtract
|
| 554 |
|
|
algorithm (timing was not examined). (This test code is version-controlled
|
| 555 |
|
|
in the same directory as this \LaTeX{} document.)
|
| 556 |
|
|
|
| 557 |
|
|
At this point I agree with the
|
| 558 |
|
|
compiler manufacturer
|
| 559 |
|
|
that there is a tradeoff between size and speed (it seems
|
| 560 |
|
|
nearly impossible to get both).
|
| 561 |
|
|
|
| 562 |
|
|
In any reimplementation of this algorithm, will probably need to choose between
|
| 563 |
|
|
size and speed. I believe there is some possibility to reduce the reimplementation
|
| 564 |
|
|
from 153 bytes, but not down to 103 bytes.
|
| 565 |
|
|
|
| 566 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 567 |
|
|
\end{document}
|
| 568 |
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
| 569 |
|
|
% $Log: cpu08div16by16a.tex,v $
|
| 570 |
|
|
% Revision 1.12 2007/05/26 18:25:46 dashley
|
| 571 |
|
|
% Edits to incorporate final thoughts (for now) and test code results.
|
| 572 |
|
|
%
|
| 573 |
|
|
% Revision 1.11 2007/05/21 16:49:40 dashley
|
| 574 |
|
|
% \bdiv functionality as suggested by Robin Fairbairns added.
|
| 575 |
|
|
%
|
| 576 |
|
|
% Revision 1.10 2007/05/18 15:53:59 dashley
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|
|
% Edits.
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%
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% Revision 1.9 2007/05/18 02:02:56 dashley
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% Edits.
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%
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% Revision 1.8 2007/05/18 02:02:10 dashley
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% Edits.
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%
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% Revision 1.7 2007/05/18 01:40:20 dashley
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% Edits.
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%
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% Revision 1.6 2007/05/18 00:15:56 dashley
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% Edits.
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%
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% Revision 1.5 2007/05/17 15:53:57 dashley
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% Edits.
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% Revision 1.4 2007/05/17 14:46:34 dashley
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% Edits.
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% Revision 1.3 2007/05/17 04:15:22 dashley
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% Revision 1.2 2007/05/17 03:50:48 dashley
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% Edits.
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%
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% Revision 1.1 2007/05/17 03:09:06 dashley
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% Initial checkin.
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| 605 |
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
Parent Directory
| 
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