urem.S 7.0 KB
Newer Older
L
Linus Torvalds 已提交
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
/* $Id: urem.S,v 1.4 1996/09/30 02:22:42 davem Exp $
 * urem.S:      This routine was taken from glibc-1.09 and is covered
 *              by the GNU Library General Public License Version 2.
 */

/* This file is generated from divrem.m4; DO NOT EDIT! */
/*
 * Division and remainder, from Appendix E of the Sparc Version 8
 * Architecture Manual, with fixes from Gordon Irlam.
 */

/*
 * Input: dividend and divisor in %o0 and %o1 respectively.
 *
 * m4 parameters:
 *  .urem	name of function to generate
 *  rem		rem=div => %o0 / %o1; rem=rem => %o0 % %o1
 *  false		false=true => signed; false=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.
 */

	.globl .urem
44
	.globl _Urem
L
Linus Torvalds 已提交
45
.urem:
46
_Urem:	/* needed for export */
L
Linus Torvalds 已提交
47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357

	! Ready to divide.  Compute size of quotient; scale comparand.
	orcc	%o1, %g0, %o5
	bne	1f
	 mov	%o0, %o3

		! Divide by zero trap.  If it returns, return 0 (about as
		! wrong as possible, but that is what SunOS does...).
		ta	ST_DIV0
		retl
		 clr	%o0

1:
	cmp	%o3, %o5			! if %o1 exceeds %o0, done
	blu	Lgot_result		! (and algorithm fails otherwise)
	 clr	%o2

	sethi	%hi(1 << (32 - 4 - 1)), %g1

	cmp	%o3, %g1
	blu	Lnot_really_big
	 clr	%o4

	! Here the dividend is >= 2**(31-N) or so.  We must be careful here,
	! as our usual N-at-a-shot divide step will cause overflow and havoc.
	! The number of bits in the result here is N*ITER+SC, where SC <= N.
	! Compute ITER in an unorthodox manner: know we need to shift V into
	! the top decade: so do not even bother to compare to R.
	1:
		cmp	%o5, %g1
		bgeu	3f
		 mov	1, %g7

		sll	%o5, 4, %o5

		b	1b
		 add	%o4, 1, %o4

	! Now compute %g7.
	2:
		addcc	%o5, %o5, %o5
		bcc	Lnot_too_big
		 add	%g7, 1, %g7

		! We get here if the %o1 overflowed while shifting.
		! This means that %o3 has the high-order bit set.
		! Restore %o5 and subtract from %o3.
		sll	%g1, 4, %g1	! high order bit
		srl	%o5, 1, %o5		! rest of %o5
		add	%o5, %g1, %o5

		b	Ldo_single_div
		 sub	%g7, 1, %g7

	Lnot_too_big:
	3:
		cmp	%o5, %o3
		blu	2b
		 nop

		be	Ldo_single_div
		 nop
	/* NB: these are commented out in the V8-Sparc manual as well */
	/* (I do not understand this) */
	! %o5 > %o3: went too far: back up 1 step
	!	srl	%o5, 1, %o5
	!	dec	%g7
	! do single-bit divide steps
	!
	! We have to be careful here.  We know that %o3 >= %o5, so we can do the
	! first divide step without thinking.  BUT, the others are conditional,
	! and are only done if %o3 >= 0.  Because both %o3 and %o5 may have the high-
	! order bit set in the first step, just falling into the regular
	! division loop will mess up the first time around.
	! So we unroll slightly...
	Ldo_single_div:
		subcc	%g7, 1, %g7
		bl	Lend_regular_divide
		 nop

		sub	%o3, %o5, %o3
		mov	1, %o2

		b	Lend_single_divloop
		 nop
	Lsingle_divloop:
		sll	%o2, 1, %o2
		bl	1f
		 srl	%o5, 1, %o5
		! %o3 >= 0
		sub	%o3, %o5, %o3
		b	2f
		 add	%o2, 1, %o2
	1:	! %o3 < 0
		add	%o3, %o5, %o3
		sub	%o2, 1, %o2
	2:
	Lend_single_divloop:
		subcc	%g7, 1, %g7
		bge	Lsingle_divloop
		 tst	%o3

		b,a	Lend_regular_divide

Lnot_really_big:
1:
	sll	%o5, 4, %o5

	cmp	%o5, %o3
	bleu	1b
	 addcc	%o4, 1, %o4

	be	Lgot_result
	 sub	%o4, 1, %o4

	tst	%o3	! set up for initial iteration
Ldivloop:
	sll	%o2, 4, %o2
		! depth 1, accumulated bits 0
	bl	L.1.16
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 2, accumulated bits 1
	bl	L.2.17
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 3, accumulated bits 3
	bl	L.3.19
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits 7
	bl	L.4.23
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (7*2+1), %o2

L.4.23:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (7*2-1), %o2

L.3.19:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits 5
	bl	L.4.21
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (5*2+1), %o2

L.4.21:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (5*2-1), %o2

L.2.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 3, accumulated bits 1
	bl	L.3.17
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits 3
	bl	L.4.19
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (3*2+1), %o2

L.4.19:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (3*2-1), %o2

L.3.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits 1
	bl	L.4.17
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (1*2+1), %o2
	
L.4.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (1*2-1), %o2

L.1.16:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 2, accumulated bits -1
	bl	L.2.15
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 3, accumulated bits -1
	bl	L.3.15
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits -1
	bl	L.4.15
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-1*2+1), %o2

L.4.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-1*2-1), %o2

L.3.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits -3
	bl	L.4.13
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-3*2+1), %o2

L.4.13:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-3*2-1), %o2

L.2.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 3, accumulated bits -3
	bl	L.3.13
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
			! depth 4, accumulated bits -5
	bl	L.4.11
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-5*2+1), %o2
	
L.4.11:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-5*2-1), %o2

L.3.13:
	! remainder is negative
	addcc	%o3,%o5,%o3
			! depth 4, accumulated bits -7
	bl	L.4.9
	 srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-7*2+1), %o2

L.4.9:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	 add	%o2, (-7*2-1), %o2

	9:
Lend_regular_divide:
	subcc	%o4, 1, %o4
	bge	Ldivloop
	 tst	%o3

	bl,a	Lgot_result
	! non-restoring fixup here (one instruction only!)
	add	%o3, %o1, %o3

Lgot_result:

	retl
	 mov %o3, %o0

	.globl	.urem_patch
.urem_patch:
	wr	%g0, 0x0, %y
	nop
	nop
	nop
	udiv	%o0, %o1, %o2
	umul	%o2, %o1, %o2
	retl
	 sub	%o0, %o2, %o0