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diff --git a/libc/sysdeps/linux/sparc/rem.S b/libc/sysdeps/linux/sparc/rem.S
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+ /* 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:
+ * .rem name of function to generate
+ * rem rem=div => %o0 / %o1; rem=rem => %o0 % %o1
+ * true true=true => signed; true=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.
+ */
+
+
+
+#include <sysdep.h>
+
+
+.global .rem;
+.align 4;
+.type .rem ,@function;
+
+.rem:
+ ! compute sign of result; if neither is negative, no problem
+ orcc %o1, %o0, %g0 ! either negative?
+ bge 2f ! no, go do the divide
+ mov %o0, %g3 ! sign of remainder matches %o0
+ tst %o1
+ bge 1f
+ tst %o0
+ ! %o1 is definitely negative; %o0 might also be negative
+ bge 2f ! if %o0 not negative...
+ sub %g0, %o1, %o1 ! in any case, make %o1 nonneg
+1: ! %o0 is negative, %o1 is nonnegative
+ sub %g0, %o0, %o0 ! make %o0 nonnegative
+2:
+
+ ! 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 0x02
+ 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, %g2
+ sll %o5, 4, %o5
+ b 1b
+ add %o4, 1, %o4
+
+ ! Now compute %g2.
+ 2: addcc %o5, %o5, %o5
+ bcc .Lnot_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 .Ldo_single_div
+ sub %g2, 1, %g2
+
+ .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 %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...
+ .Ldo_single_div:
+ subcc %g2, 1, %g2
+ 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 %g2, 1, %g2
+ 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 .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:
+.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:
+ ! check to see if answer should be < 0
+ tst %g3
+ bl,a 1f
+ sub %g0, %o3, %o3
+1:
+ retl
+ mov %o3, %o0
+
+END(.rem)