kernel-ark/arch/parisc/lib/milli/div_const.S
Kyle McMartin efb80e7e09 [PARISC] import necessary bits of libgcc.a
Currently we're hacking libs-y to include libgcc.a, but this has
unforeseen consequences since the userspace libgcc is linked with fpregs
enabled. We need the kernel to stop using fpregs in an uncontrolled manner
to implement lazy fpu state saves.

Signed-off-by: Kyle McMartin <kyle@mcmartin.ca>
2007-10-18 00:58:49 -07:00

683 lines
18 KiB
ArmAsm

/* 32 and 64-bit millicode, original author Hewlett-Packard
adapted for gcc by Paul Bame <bame@debian.org>
and Alan Modra <alan@linuxcare.com.au>.
Copyright 2001, 2002, 2003 Free Software Foundation, Inc.
This file is part of GCC and is released under the terms of
of the GNU General Public License as published by the Free Software
Foundation; either version 2, or (at your option) any later version.
See the file COPYING in the top-level GCC source directory for a copy
of the license. */
#include "milli.h"
#ifdef L_div_const
/* ROUTINE: $$divI_2
. $$divI_3 $$divU_3
. $$divI_4
. $$divI_5 $$divU_5
. $$divI_6 $$divU_6
. $$divI_7 $$divU_7
. $$divI_8
. $$divI_9 $$divU_9
. $$divI_10 $$divU_10
.
. $$divI_12 $$divU_12
.
. $$divI_14 $$divU_14
. $$divI_15 $$divU_15
. $$divI_16
. $$divI_17 $$divU_17
.
. Divide by selected constants for single precision binary integers.
INPUT REGISTERS:
. arg0 == dividend
. mrp == return pc
. sr0 == return space when called externally
OUTPUT REGISTERS:
. arg0 = undefined
. arg1 = undefined
. ret1 = quotient
OTHER REGISTERS AFFECTED:
. r1 = undefined
SIDE EFFECTS:
. Causes a trap under the following conditions: NONE
. Changes memory at the following places: NONE
PERMISSIBLE CONTEXT:
. Unwindable.
. Does not create a stack frame.
. Suitable for internal or external millicode.
. Assumes the special millicode register conventions.
DISCUSSION:
. Calls other millicode routines using mrp: NONE
. Calls other millicode routines: NONE */
/* TRUNCATED DIVISION BY SMALL INTEGERS
We are interested in q(x) = floor(x/y), where x >= 0 and y > 0
(with y fixed).
Let a = floor(z/y), for some choice of z. Note that z will be
chosen so that division by z is cheap.
Let r be the remainder(z/y). In other words, r = z - ay.
Now, our method is to choose a value for b such that
q'(x) = floor((ax+b)/z)
is equal to q(x) over as large a range of x as possible. If the
two are equal over a sufficiently large range, and if it is easy to
form the product (ax), and it is easy to divide by z, then we can
perform the division much faster than the general division algorithm.
So, we want the following to be true:
. For x in the following range:
.
. ky <= x < (k+1)y
.
. implies that
.
. k <= (ax+b)/z < (k+1)
We want to determine b such that this is true for all k in the
range {0..K} for some maximum K.
Since (ax+b) is an increasing function of x, we can take each
bound separately to determine the "best" value for b.
(ax+b)/z < (k+1) implies
(a((k+1)y-1)+b < (k+1)z implies
b < a + (k+1)(z-ay) implies
b < a + (k+1)r
This needs to be true for all k in the range {0..K}. In
particular, it is true for k = 0 and this leads to a maximum
acceptable value for b.
b < a+r or b <= a+r-1
Taking the other bound, we have
k <= (ax+b)/z implies
k <= (aky+b)/z implies
k(z-ay) <= b implies
kr <= b
Clearly, the largest range for k will be achieved by maximizing b,
when r is not zero. When r is zero, then the simplest choice for b
is 0. When r is not 0, set
. b = a+r-1
Now, by construction, q'(x) = floor((ax+b)/z) = q(x) = floor(x/y)
for all x in the range:
. 0 <= x < (K+1)y
We need to determine what K is. Of our two bounds,
. b < a+(k+1)r is satisfied for all k >= 0, by construction.
The other bound is
. kr <= b
This is always true if r = 0. If r is not 0 (the usual case), then
K = floor((a+r-1)/r), is the maximum value for k.
Therefore, the formula q'(x) = floor((ax+b)/z) yields the correct
answer for q(x) = floor(x/y) when x is in the range
(0,(K+1)y-1) K = floor((a+r-1)/r)
To be most useful, we want (K+1)y-1 = (max x) >= 2**32-1 so that
the formula for q'(x) yields the correct value of q(x) for all x
representable by a single word in HPPA.
We are also constrained in that computing the product (ax), adding
b, and dividing by z must all be done quickly, otherwise we will be
better off going through the general algorithm using the DS
instruction, which uses approximately 70 cycles.
For each y, there is a choice of z which satisfies the constraints
for (K+1)y >= 2**32. We may not, however, be able to satisfy the
timing constraints for arbitrary y. It seems that z being equal to
a power of 2 or a power of 2 minus 1 is as good as we can do, since
it minimizes the time to do division by z. We want the choice of z
to also result in a value for (a) that minimizes the computation of
the product (ax). This is best achieved if (a) has a regular bit
pattern (so the multiplication can be done with shifts and adds).
The value of (a) also needs to be less than 2**32 so the product is
always guaranteed to fit in 2 words.
In actual practice, the following should be done:
1) For negative x, you should take the absolute value and remember
. the fact so that the result can be negated. This obviously does
. not apply in the unsigned case.
2) For even y, you should factor out the power of 2 that divides y
. and divide x by it. You can then proceed by dividing by the
. odd factor of y.
Here is a table of some odd values of y, and corresponding choices
for z which are "good".
y z r a (hex) max x (hex)
3 2**32 1 55555555 100000001
5 2**32 1 33333333 100000003
7 2**24-1 0 249249 (infinite)
9 2**24-1 0 1c71c7 (infinite)
11 2**20-1 0 1745d (infinite)
13 2**24-1 0 13b13b (infinite)
15 2**32 1 11111111 10000000d
17 2**32 1 f0f0f0f 10000000f
If r is 1, then b = a+r-1 = a. This simplifies the computation
of (ax+b), since you can compute (x+1)(a) instead. If r is 0,
then b = 0 is ok to use which simplifies (ax+b).
The bit patterns for 55555555, 33333333, and 11111111 are obviously
very regular. The bit patterns for the other values of a above are:
y (hex) (binary)
7 249249 001001001001001001001001 << regular >>
9 1c71c7 000111000111000111000111 << regular >>
11 1745d 000000010111010001011101 << irregular >>
13 13b13b 000100111011000100111011 << irregular >>
The bit patterns for (a) corresponding to (y) of 11 and 13 may be
too irregular to warrant using this method.
When z is a power of 2 minus 1, then the division by z is slightly
more complicated, involving an iterative solution.
The code presented here solves division by 1 through 17, except for
11 and 13. There are algorithms for both signed and unsigned
quantities given.
TIMINGS (cycles)
divisor positive negative unsigned
. 1 2 2 2
. 2 4 4 2
. 3 19 21 19
. 4 4 4 2
. 5 18 22 19
. 6 19 22 19
. 8 4 4 2
. 10 18 19 17
. 12 18 20 18
. 15 16 18 16
. 16 4 4 2
. 17 16 18 16
Now, the algorithm for 7, 9, and 14 is an iterative one. That is,
a loop body is executed until the tentative quotient is 0. The
number of times the loop body is executed varies depending on the
dividend, but is never more than two times. If the dividend is
less than the divisor, then the loop body is not executed at all.
Each iteration adds 4 cycles to the timings.
divisor positive negative unsigned
. 7 19+4n 20+4n 20+4n n = number of iterations
. 9 21+4n 22+4n 21+4n
. 14 21+4n 22+4n 20+4n
To give an idea of how the number of iterations varies, here is a
table of dividend versus number of iterations when dividing by 7.
smallest largest required
dividend dividend iterations
. 0 6 0
. 7 0x6ffffff 1
0x1000006 0xffffffff 2
There is some overlap in the range of numbers requiring 1 and 2
iterations. */
RDEFINE(t2,r1)
RDEFINE(x2,arg0) /* r26 */
RDEFINE(t1,arg1) /* r25 */
RDEFINE(x1,ret1) /* r29 */
SUBSPA_MILLI_DIV
ATTR_MILLI
.proc
.callinfo millicode
.entry
/* NONE of these routines require a stack frame
ALL of these routines are unwindable from millicode */
GSYM($$divide_by_constant)
.export $$divide_by_constant,millicode
/* Provides a "nice" label for the code covered by the unwind descriptor
for things like gprof. */
/* DIVISION BY 2 (shift by 1) */
GSYM($$divI_2)
.export $$divI_2,millicode
comclr,>= arg0,0,0
addi 1,arg0,arg0
MILLIRET
extrs arg0,30,31,ret1
/* DIVISION BY 4 (shift by 2) */
GSYM($$divI_4)
.export $$divI_4,millicode
comclr,>= arg0,0,0
addi 3,arg0,arg0
MILLIRET
extrs arg0,29,30,ret1
/* DIVISION BY 8 (shift by 3) */
GSYM($$divI_8)
.export $$divI_8,millicode
comclr,>= arg0,0,0
addi 7,arg0,arg0
MILLIRET
extrs arg0,28,29,ret1
/* DIVISION BY 16 (shift by 4) */
GSYM($$divI_16)
.export $$divI_16,millicode
comclr,>= arg0,0,0
addi 15,arg0,arg0
MILLIRET
extrs arg0,27,28,ret1
/****************************************************************************
*
* DIVISION BY DIVISORS OF FFFFFFFF, and powers of 2 times these
*
* includes 3,5,15,17 and also 6,10,12
*
****************************************************************************/
/* DIVISION BY 3 (use z = 2**32; a = 55555555) */
GSYM($$divI_3)
.export $$divI_3,millicode
comb,<,N x2,0,LREF(neg3)
addi 1,x2,x2 /* this cannot overflow */
extru x2,1,2,x1 /* multiply by 5 to get started */
sh2add x2,x2,x2
b LREF(pos)
addc x1,0,x1
LSYM(neg3)
subi 1,x2,x2 /* this cannot overflow */
extru x2,1,2,x1 /* multiply by 5 to get started */
sh2add x2,x2,x2
b LREF(neg)
addc x1,0,x1
GSYM($$divU_3)
.export $$divU_3,millicode
addi 1,x2,x2 /* this CAN overflow */
addc 0,0,x1
shd x1,x2,30,t1 /* multiply by 5 to get started */
sh2add x2,x2,x2
b LREF(pos)
addc x1,t1,x1
/* DIVISION BY 5 (use z = 2**32; a = 33333333) */
GSYM($$divI_5)
.export $$divI_5,millicode
comb,<,N x2,0,LREF(neg5)
addi 3,x2,t1 /* this cannot overflow */
sh1add x2,t1,x2 /* multiply by 3 to get started */
b LREF(pos)
addc 0,0,x1
LSYM(neg5)
sub 0,x2,x2 /* negate x2 */
addi 1,x2,x2 /* this cannot overflow */
shd 0,x2,31,x1 /* get top bit (can be 1) */
sh1add x2,x2,x2 /* multiply by 3 to get started */
b LREF(neg)
addc x1,0,x1
GSYM($$divU_5)
.export $$divU_5,millicode
addi 1,x2,x2 /* this CAN overflow */
addc 0,0,x1
shd x1,x2,31,t1 /* multiply by 3 to get started */
sh1add x2,x2,x2
b LREF(pos)
addc t1,x1,x1
/* DIVISION BY 6 (shift to divide by 2 then divide by 3) */
GSYM($$divI_6)
.export $$divI_6,millicode
comb,<,N x2,0,LREF(neg6)
extru x2,30,31,x2 /* divide by 2 */
addi 5,x2,t1 /* compute 5*(x2+1) = 5*x2+5 */
sh2add x2,t1,x2 /* multiply by 5 to get started */
b LREF(pos)
addc 0,0,x1
LSYM(neg6)
subi 2,x2,x2 /* negate, divide by 2, and add 1 */
/* negation and adding 1 are done */
/* at the same time by the SUBI */
extru x2,30,31,x2
shd 0,x2,30,x1
sh2add x2,x2,x2 /* multiply by 5 to get started */
b LREF(neg)
addc x1,0,x1
GSYM($$divU_6)
.export $$divU_6,millicode
extru x2,30,31,x2 /* divide by 2 */
addi 1,x2,x2 /* cannot carry */
shd 0,x2,30,x1 /* multiply by 5 to get started */
sh2add x2,x2,x2
b LREF(pos)
addc x1,0,x1
/* DIVISION BY 10 (shift to divide by 2 then divide by 5) */
GSYM($$divU_10)
.export $$divU_10,millicode
extru x2,30,31,x2 /* divide by 2 */
addi 3,x2,t1 /* compute 3*(x2+1) = (3*x2)+3 */
sh1add x2,t1,x2 /* multiply by 3 to get started */
addc 0,0,x1
LSYM(pos)
shd x1,x2,28,t1 /* multiply by 0x11 */
shd x2,0,28,t2
add x2,t2,x2
addc x1,t1,x1
LSYM(pos_for_17)
shd x1,x2,24,t1 /* multiply by 0x101 */
shd x2,0,24,t2
add x2,t2,x2
addc x1,t1,x1
shd x1,x2,16,t1 /* multiply by 0x10001 */
shd x2,0,16,t2
add x2,t2,x2
MILLIRET
addc x1,t1,x1
GSYM($$divI_10)
.export $$divI_10,millicode
comb,< x2,0,LREF(neg10)
copy 0,x1
extru x2,30,31,x2 /* divide by 2 */
addib,TR 1,x2,LREF(pos) /* add 1 (cannot overflow) */
sh1add x2,x2,x2 /* multiply by 3 to get started */
LSYM(neg10)
subi 2,x2,x2 /* negate, divide by 2, and add 1 */
/* negation and adding 1 are done */
/* at the same time by the SUBI */
extru x2,30,31,x2
sh1add x2,x2,x2 /* multiply by 3 to get started */
LSYM(neg)
shd x1,x2,28,t1 /* multiply by 0x11 */
shd x2,0,28,t2
add x2,t2,x2
addc x1,t1,x1
LSYM(neg_for_17)
shd x1,x2,24,t1 /* multiply by 0x101 */
shd x2,0,24,t2
add x2,t2,x2
addc x1,t1,x1
shd x1,x2,16,t1 /* multiply by 0x10001 */
shd x2,0,16,t2
add x2,t2,x2
addc x1,t1,x1
MILLIRET
sub 0,x1,x1
/* DIVISION BY 12 (shift to divide by 4 then divide by 3) */
GSYM($$divI_12)
.export $$divI_12,millicode
comb,< x2,0,LREF(neg12)
copy 0,x1
extru x2,29,30,x2 /* divide by 4 */
addib,tr 1,x2,LREF(pos) /* compute 5*(x2+1) = 5*x2+5 */
sh2add x2,x2,x2 /* multiply by 5 to get started */
LSYM(neg12)
subi 4,x2,x2 /* negate, divide by 4, and add 1 */
/* negation and adding 1 are done */
/* at the same time by the SUBI */
extru x2,29,30,x2
b LREF(neg)
sh2add x2,x2,x2 /* multiply by 5 to get started */
GSYM($$divU_12)
.export $$divU_12,millicode
extru x2,29,30,x2 /* divide by 4 */
addi 5,x2,t1 /* cannot carry */
sh2add x2,t1,x2 /* multiply by 5 to get started */
b LREF(pos)
addc 0,0,x1
/* DIVISION BY 15 (use z = 2**32; a = 11111111) */
GSYM($$divI_15)
.export $$divI_15,millicode
comb,< x2,0,LREF(neg15)
copy 0,x1
addib,tr 1,x2,LREF(pos)+4
shd x1,x2,28,t1
LSYM(neg15)
b LREF(neg)
subi 1,x2,x2
GSYM($$divU_15)
.export $$divU_15,millicode
addi 1,x2,x2 /* this CAN overflow */
b LREF(pos)
addc 0,0,x1
/* DIVISION BY 17 (use z = 2**32; a = f0f0f0f) */
GSYM($$divI_17)
.export $$divI_17,millicode
comb,<,n x2,0,LREF(neg17)
addi 1,x2,x2 /* this cannot overflow */
shd 0,x2,28,t1 /* multiply by 0xf to get started */
shd x2,0,28,t2
sub t2,x2,x2
b LREF(pos_for_17)
subb t1,0,x1
LSYM(neg17)
subi 1,x2,x2 /* this cannot overflow */
shd 0,x2,28,t1 /* multiply by 0xf to get started */
shd x2,0,28,t2
sub t2,x2,x2
b LREF(neg_for_17)
subb t1,0,x1
GSYM($$divU_17)
.export $$divU_17,millicode
addi 1,x2,x2 /* this CAN overflow */
addc 0,0,x1
shd x1,x2,28,t1 /* multiply by 0xf to get started */
LSYM(u17)
shd x2,0,28,t2
sub t2,x2,x2
b LREF(pos_for_17)
subb t1,x1,x1
/* DIVISION BY DIVISORS OF FFFFFF, and powers of 2 times these
includes 7,9 and also 14
z = 2**24-1
r = z mod x = 0
so choose b = 0
Also, in order to divide by z = 2**24-1, we approximate by dividing
by (z+1) = 2**24 (which is easy), and then correcting.
(ax) = (z+1)q' + r
. = zq' + (q'+r)
So to compute (ax)/z, compute q' = (ax)/(z+1) and r = (ax) mod (z+1)
Then the true remainder of (ax)/z is (q'+r). Repeat the process
with this new remainder, adding the tentative quotients together,
until a tentative quotient is 0 (and then we are done). There is
one last correction to be done. It is possible that (q'+r) = z.
If so, then (q'+r)/(z+1) = 0 and it looks like we are done. But,
in fact, we need to add 1 more to the quotient. Now, it turns
out that this happens if and only if the original value x is
an exact multiple of y. So, to avoid a three instruction test at
the end, instead use 1 instruction to add 1 to x at the beginning. */
/* DIVISION BY 7 (use z = 2**24-1; a = 249249) */
GSYM($$divI_7)
.export $$divI_7,millicode
comb,<,n x2,0,LREF(neg7)
LSYM(7)
addi 1,x2,x2 /* cannot overflow */
shd 0,x2,29,x1
sh3add x2,x2,x2
addc x1,0,x1
LSYM(pos7)
shd x1,x2,26,t1
shd x2,0,26,t2
add x2,t2,x2
addc x1,t1,x1
shd x1,x2,20,t1
shd x2,0,20,t2
add x2,t2,x2
addc x1,t1,t1
/* computed <t1,x2>. Now divide it by (2**24 - 1) */
copy 0,x1
shd,= t1,x2,24,t1 /* tentative quotient */
LSYM(1)
addb,tr t1,x1,LREF(2) /* add to previous quotient */
extru x2,31,24,x2 /* new remainder (unadjusted) */
MILLIRETN
LSYM(2)
addb,tr t1,x2,LREF(1) /* adjust remainder */
extru,= x2,7,8,t1 /* new quotient */
LSYM(neg7)
subi 1,x2,x2 /* negate x2 and add 1 */
LSYM(8)
shd 0,x2,29,x1
sh3add x2,x2,x2
addc x1,0,x1
LSYM(neg7_shift)
shd x1,x2,26,t1
shd x2,0,26,t2
add x2,t2,x2
addc x1,t1,x1
shd x1,x2,20,t1
shd x2,0,20,t2
add x2,t2,x2
addc x1,t1,t1
/* computed <t1,x2>. Now divide it by (2**24 - 1) */
copy 0,x1
shd,= t1,x2,24,t1 /* tentative quotient */
LSYM(3)
addb,tr t1,x1,LREF(4) /* add to previous quotient */
extru x2,31,24,x2 /* new remainder (unadjusted) */
MILLIRET
sub 0,x1,x1 /* negate result */
LSYM(4)
addb,tr t1,x2,LREF(3) /* adjust remainder */
extru,= x2,7,8,t1 /* new quotient */
GSYM($$divU_7)
.export $$divU_7,millicode
addi 1,x2,x2 /* can carry */
addc 0,0,x1
shd x1,x2,29,t1
sh3add x2,x2,x2
b LREF(pos7)
addc t1,x1,x1
/* DIVISION BY 9 (use z = 2**24-1; a = 1c71c7) */
GSYM($$divI_9)
.export $$divI_9,millicode
comb,<,n x2,0,LREF(neg9)
addi 1,x2,x2 /* cannot overflow */
shd 0,x2,29,t1
shd x2,0,29,t2
sub t2,x2,x2
b LREF(pos7)
subb t1,0,x1
LSYM(neg9)
subi 1,x2,x2 /* negate and add 1 */
shd 0,x2,29,t1
shd x2,0,29,t2
sub t2,x2,x2
b LREF(neg7_shift)
subb t1,0,x1
GSYM($$divU_9)
.export $$divU_9,millicode
addi 1,x2,x2 /* can carry */
addc 0,0,x1
shd x1,x2,29,t1
shd x2,0,29,t2
sub t2,x2,x2
b LREF(pos7)
subb t1,x1,x1
/* DIVISION BY 14 (shift to divide by 2 then divide by 7) */
GSYM($$divI_14)
.export $$divI_14,millicode
comb,<,n x2,0,LREF(neg14)
GSYM($$divU_14)
.export $$divU_14,millicode
b LREF(7) /* go to 7 case */
extru x2,30,31,x2 /* divide by 2 */
LSYM(neg14)
subi 2,x2,x2 /* negate (and add 2) */
b LREF(8)
extru x2,30,31,x2 /* divide by 2 */
.exit
.procend
.end
#endif