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M4 Source File | 1994-10-21 | 5.8 KB | 235 lines

/* * Division and remainder, from Appendix E of the Sparc Version 8 * Architecture Manual, with fixes from Gordon Irlam. */ /* * Input: dividend and divisor in %o0 and %o1 respectively. * * m4 parameters: * NAME name of function to generate * OP OP=div => %o0 / %o1; OP=rem => %o0 % %o1 * S S=true => signed; S=false => unsigned * * Algorithm parameters: * N how many bits per iteration we try to get (4) * WORDSIZE total number of bits (32) * * Derived constants: * TOPBITS number of bits in the top `decade' of a number * * Important variables: * Q the partial quotient under development (initially 0) * R the remainder so far, initially the dividend * ITER number of main division loop iterations required; * equal to ceil(log2(quotient) / N). Note that this * is the log base (2^N) of the quotient. * V the current comparand, initially divisor*2^(ITER*N-1) * * Cost: * Current estimate for non-large dividend is * ceil(log2(quotient) / N) * (10 + 7N/2) + C * A large dividend is one greater than 2^(31-TOPBITS) and takes a * different path, as the upper bits of the quotient must be developed * one bit at a time. */ define(N, `4')dnl define(WORDSIZE, `32')dnl define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))dnl dnl define(dividend, `%o0')dnl define(divisor, `%o1')dnl define(Q, `%o2')dnl define(R, `%o3')dnl define(ITER, `%o4')dnl define(V, `%o5')dnl dnl dnl m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d define(T, `%g1')dnl define(SC, `%g7')dnl ifelse(S, `true', `define(SIGN, `%g6')')dnl dnl dnl This is the recursive definition for developing quotient digits. dnl dnl Parameters: dnl $1 the current depth, 1 <= $1 <= N dnl $2 the current accumulation of quotient bits dnl N max depth dnl dnl We add a new bit to $2 and either recurse or insert the bits in dnl the quotient. R, Q, and V are inputs and outputs as defined above; dnl the condition codes are expected to reflect the input R, and are dnl modified to reflect the output R. dnl define(DEVELOP_QUOTIENT_BITS, ` ! depth $1, accumulated bits $2 bl L.$1.eval(2**N+$2) srl V,1,V ! remainder is positive subcc R,V,R ifelse($1, N, ` b 9f add Q, ($2*2+1), Q ', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')') L.$1.eval(2**N+$2): ! remainder is negative addcc R,V,R ifelse($1, N, ` b 9f add Q, ($2*2-1), Q ', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')') ifelse($1, 1, `9:')')dnl #include "DEFS.h" #ifdef __svr4__ #include <sys/trap.h> #else #include <machine/trap.h> #endif FUNC(NAME) ifelse(S, `true', ` ! compute sign of result; if neither is negative, no problem orcc divisor, dividend, %g0 ! either negative? bge 2f ! no, go do the divide ifelse(OP, `div', ` xor divisor, dividend, SIGN ! compute sign in any case', ` mov dividend, SIGN ! sign of remainder matches dividend') tst divisor bge 1f tst dividend ! divisor is definitely negative; dividend might also be negative bge 2f ! if dividend not negative... sub %g0, divisor, divisor ! in any case, make divisor nonneg 1: ! dividend is negative, divisor is nonnegative sub %g0, dividend, dividend ! make dividend nonnegative 2: ') ! Ready to divide. Compute size of quotient; scale comparand. orcc divisor, %g0, V bne 1f mov dividend, R ! Divide by zero trap. If it returns, return 0 (about as ! wrong as possible, but that is what SunOS does...). ta ST_DIV0 retl clr %o0 1: cmp R, V ! if divisor exceeds dividend, done blu Lgot_result ! (and algorithm fails otherwise) clr Q sethi %hi(1 << (WORDSIZE - TOPBITS - 1)), T cmp R, T blu Lnot_really_big clr ITER ! `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 V, T bgeu 3f mov 1, SC sll V, N, V b 1b add ITER, 1, ITER ! Now compute SC. 2: addcc V, V, V bcc Lnot_too_big add SC, 1, SC ! We get here if the divisor overflowed while shifting. ! This means that R has the high-order bit set. ! Restore V and subtract from R. sll T, TOPBITS, T ! high order bit srl V, 1, V ! rest of V add V, T, V b Ldo_single_div sub SC, 1, SC Lnot_too_big: 3: cmp V, R blu 2b nop be Ldo_single_div nop /* NB: these are commented out in the V8-Sparc manual as well */ /* (I do not understand this) */ ! V > R: went too far: back up 1 step ! srl V, 1, V ! dec SC ! do single-bit divide steps ! ! We have to be careful here. We know that R >= V, so we can do the ! first divide step without thinking. BUT, the others are conditional, ! and are only done if R >= 0. Because both R and V may have the high- ! order bit set in the first step, just falling into the regular ! division loop will mess up the first time around. ! So we unroll slightly... Ldo_single_div: subcc SC, 1, SC bl Lend_regular_divide nop sub R, V, R mov 1, Q b Lend_single_divloop nop Lsingle_divloop: sll Q, 1, Q bl 1f srl V, 1, V ! R >= 0 sub R, V, R b 2f add Q, 1, Q 1: ! R < 0 add R, V, R sub Q, 1, Q 2: Lend_single_divloop: subcc SC, 1, SC bge Lsingle_divloop tst R b,a Lend_regular_divide Lnot_really_big: 1: sll V, N, V cmp V, R bleu 1b addcc ITER, 1, ITER be Lgot_result sub ITER, 1, ITER tst R ! set up for initial iteration Ldivloop: sll Q, N, Q DEVELOP_QUOTIENT_BITS(1, 0) Lend_regular_divide: subcc ITER, 1, ITER bge Ldivloop tst R bl,a Lgot_result ! non-restoring fixup here (one instruction only!) ifelse(OP, `div', ` sub Q, 1, Q ', ` add R, divisor, R ') Lgot_result: ifelse(S, `true', ` ! check to see if answer should be < 0 tst SIGN bl,a 1f ifelse(OP, `div', `sub %g0, Q, Q', `sub %g0, R, R') 1:') retl ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')