efb80e7e09
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>
683 lines
18 KiB
ArmAsm
683 lines
18 KiB
ArmAsm
/* 32 and 64-bit millicode, original author Hewlett-Packard
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adapted for gcc by Paul Bame <bame@debian.org>
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and Alan Modra <alan@linuxcare.com.au>.
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Copyright 2001, 2002, 2003 Free Software Foundation, Inc.
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This file is part of GCC and is released under the terms of
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of the GNU General Public License as published by the Free Software
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Foundation; either version 2, or (at your option) any later version.
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See the file COPYING in the top-level GCC source directory for a copy
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of the license. */
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#include "milli.h"
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#ifdef L_div_const
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/* ROUTINE: $$divI_2
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. $$divI_3 $$divU_3
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. $$divI_4
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. $$divI_5 $$divU_5
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. $$divI_6 $$divU_6
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. $$divI_7 $$divU_7
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. $$divI_8
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. $$divI_9 $$divU_9
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. $$divI_10 $$divU_10
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.
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. $$divI_12 $$divU_12
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.
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. $$divI_14 $$divU_14
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. $$divI_15 $$divU_15
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. $$divI_16
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. $$divI_17 $$divU_17
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.
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. Divide by selected constants for single precision binary integers.
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INPUT REGISTERS:
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. arg0 == dividend
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. mrp == return pc
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. sr0 == return space when called externally
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OUTPUT REGISTERS:
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. arg0 = undefined
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. arg1 = undefined
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. ret1 = quotient
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OTHER REGISTERS AFFECTED:
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. r1 = undefined
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SIDE EFFECTS:
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. Causes a trap under the following conditions: NONE
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. Changes memory at the following places: NONE
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PERMISSIBLE CONTEXT:
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. Unwindable.
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. Does not create a stack frame.
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. Suitable for internal or external millicode.
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. Assumes the special millicode register conventions.
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DISCUSSION:
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. Calls other millicode routines using mrp: NONE
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. Calls other millicode routines: NONE */
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/* TRUNCATED DIVISION BY SMALL INTEGERS
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We are interested in q(x) = floor(x/y), where x >= 0 and y > 0
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(with y fixed).
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Let a = floor(z/y), for some choice of z. Note that z will be
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chosen so that division by z is cheap.
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Let r be the remainder(z/y). In other words, r = z - ay.
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Now, our method is to choose a value for b such that
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q'(x) = floor((ax+b)/z)
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is equal to q(x) over as large a range of x as possible. If the
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two are equal over a sufficiently large range, and if it is easy to
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form the product (ax), and it is easy to divide by z, then we can
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perform the division much faster than the general division algorithm.
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So, we want the following to be true:
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. For x in the following range:
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.
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. ky <= x < (k+1)y
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.
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. implies that
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.
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. k <= (ax+b)/z < (k+1)
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We want to determine b such that this is true for all k in the
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range {0..K} for some maximum K.
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Since (ax+b) is an increasing function of x, we can take each
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bound separately to determine the "best" value for b.
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(ax+b)/z < (k+1) implies
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(a((k+1)y-1)+b < (k+1)z implies
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b < a + (k+1)(z-ay) implies
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b < a + (k+1)r
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This needs to be true for all k in the range {0..K}. In
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particular, it is true for k = 0 and this leads to a maximum
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acceptable value for b.
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b < a+r or b <= a+r-1
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Taking the other bound, we have
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k <= (ax+b)/z implies
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k <= (aky+b)/z implies
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k(z-ay) <= b implies
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kr <= b
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Clearly, the largest range for k will be achieved by maximizing b,
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when r is not zero. When r is zero, then the simplest choice for b
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is 0. When r is not 0, set
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. b = a+r-1
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Now, by construction, q'(x) = floor((ax+b)/z) = q(x) = floor(x/y)
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for all x in the range:
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. 0 <= x < (K+1)y
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We need to determine what K is. Of our two bounds,
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. b < a+(k+1)r is satisfied for all k >= 0, by construction.
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The other bound is
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. kr <= b
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This is always true if r = 0. If r is not 0 (the usual case), then
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K = floor((a+r-1)/r), is the maximum value for k.
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Therefore, the formula q'(x) = floor((ax+b)/z) yields the correct
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answer for q(x) = floor(x/y) when x is in the range
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(0,(K+1)y-1) K = floor((a+r-1)/r)
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To be most useful, we want (K+1)y-1 = (max x) >= 2**32-1 so that
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the formula for q'(x) yields the correct value of q(x) for all x
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representable by a single word in HPPA.
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We are also constrained in that computing the product (ax), adding
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b, and dividing by z must all be done quickly, otherwise we will be
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better off going through the general algorithm using the DS
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instruction, which uses approximately 70 cycles.
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For each y, there is a choice of z which satisfies the constraints
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for (K+1)y >= 2**32. We may not, however, be able to satisfy the
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timing constraints for arbitrary y. It seems that z being equal to
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a power of 2 or a power of 2 minus 1 is as good as we can do, since
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it minimizes the time to do division by z. We want the choice of z
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to also result in a value for (a) that minimizes the computation of
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the product (ax). This is best achieved if (a) has a regular bit
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pattern (so the multiplication can be done with shifts and adds).
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The value of (a) also needs to be less than 2**32 so the product is
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always guaranteed to fit in 2 words.
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In actual practice, the following should be done:
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1) For negative x, you should take the absolute value and remember
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. the fact so that the result can be negated. This obviously does
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. not apply in the unsigned case.
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2) For even y, you should factor out the power of 2 that divides y
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. and divide x by it. You can then proceed by dividing by the
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. odd factor of y.
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Here is a table of some odd values of y, and corresponding choices
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for z which are "good".
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y z r a (hex) max x (hex)
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3 2**32 1 55555555 100000001
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5 2**32 1 33333333 100000003
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7 2**24-1 0 249249 (infinite)
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9 2**24-1 0 1c71c7 (infinite)
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11 2**20-1 0 1745d (infinite)
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13 2**24-1 0 13b13b (infinite)
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15 2**32 1 11111111 10000000d
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17 2**32 1 f0f0f0f 10000000f
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If r is 1, then b = a+r-1 = a. This simplifies the computation
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of (ax+b), since you can compute (x+1)(a) instead. If r is 0,
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then b = 0 is ok to use which simplifies (ax+b).
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The bit patterns for 55555555, 33333333, and 11111111 are obviously
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very regular. The bit patterns for the other values of a above are:
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y (hex) (binary)
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7 249249 001001001001001001001001 << regular >>
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9 1c71c7 000111000111000111000111 << regular >>
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11 1745d 000000010111010001011101 << irregular >>
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13 13b13b 000100111011000100111011 << irregular >>
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The bit patterns for (a) corresponding to (y) of 11 and 13 may be
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too irregular to warrant using this method.
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When z is a power of 2 minus 1, then the division by z is slightly
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more complicated, involving an iterative solution.
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The code presented here solves division by 1 through 17, except for
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11 and 13. There are algorithms for both signed and unsigned
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quantities given.
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TIMINGS (cycles)
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divisor positive negative unsigned
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. 1 2 2 2
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. 2 4 4 2
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. 3 19 21 19
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. 4 4 4 2
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. 5 18 22 19
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. 6 19 22 19
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. 8 4 4 2
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. 10 18 19 17
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. 12 18 20 18
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. 15 16 18 16
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. 16 4 4 2
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. 17 16 18 16
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Now, the algorithm for 7, 9, and 14 is an iterative one. That is,
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a loop body is executed until the tentative quotient is 0. The
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number of times the loop body is executed varies depending on the
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dividend, but is never more than two times. If the dividend is
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less than the divisor, then the loop body is not executed at all.
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Each iteration adds 4 cycles to the timings.
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divisor positive negative unsigned
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. 7 19+4n 20+4n 20+4n n = number of iterations
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. 9 21+4n 22+4n 21+4n
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. 14 21+4n 22+4n 20+4n
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To give an idea of how the number of iterations varies, here is a
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table of dividend versus number of iterations when dividing by 7.
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smallest largest required
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dividend dividend iterations
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. 0 6 0
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. 7 0x6ffffff 1
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0x1000006 0xffffffff 2
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There is some overlap in the range of numbers requiring 1 and 2
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iterations. */
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RDEFINE(t2,r1)
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RDEFINE(x2,arg0) /* r26 */
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RDEFINE(t1,arg1) /* r25 */
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RDEFINE(x1,ret1) /* r29 */
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SUBSPA_MILLI_DIV
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ATTR_MILLI
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.proc
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.callinfo millicode
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.entry
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/* NONE of these routines require a stack frame
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ALL of these routines are unwindable from millicode */
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GSYM($$divide_by_constant)
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.export $$divide_by_constant,millicode
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/* Provides a "nice" label for the code covered by the unwind descriptor
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for things like gprof. */
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/* DIVISION BY 2 (shift by 1) */
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GSYM($$divI_2)
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.export $$divI_2,millicode
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comclr,>= arg0,0,0
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addi 1,arg0,arg0
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MILLIRET
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extrs arg0,30,31,ret1
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/* DIVISION BY 4 (shift by 2) */
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GSYM($$divI_4)
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.export $$divI_4,millicode
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comclr,>= arg0,0,0
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addi 3,arg0,arg0
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MILLIRET
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extrs arg0,29,30,ret1
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/* DIVISION BY 8 (shift by 3) */
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GSYM($$divI_8)
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.export $$divI_8,millicode
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comclr,>= arg0,0,0
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addi 7,arg0,arg0
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MILLIRET
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extrs arg0,28,29,ret1
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/* DIVISION BY 16 (shift by 4) */
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GSYM($$divI_16)
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.export $$divI_16,millicode
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comclr,>= arg0,0,0
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addi 15,arg0,arg0
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MILLIRET
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extrs arg0,27,28,ret1
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/****************************************************************************
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*
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* DIVISION BY DIVISORS OF FFFFFFFF, and powers of 2 times these
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*
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* includes 3,5,15,17 and also 6,10,12
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*
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****************************************************************************/
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/* DIVISION BY 3 (use z = 2**32; a = 55555555) */
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GSYM($$divI_3)
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.export $$divI_3,millicode
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comb,<,N x2,0,LREF(neg3)
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addi 1,x2,x2 /* this cannot overflow */
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extru x2,1,2,x1 /* multiply by 5 to get started */
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sh2add x2,x2,x2
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b LREF(pos)
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addc x1,0,x1
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LSYM(neg3)
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subi 1,x2,x2 /* this cannot overflow */
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extru x2,1,2,x1 /* multiply by 5 to get started */
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sh2add x2,x2,x2
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b LREF(neg)
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addc x1,0,x1
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GSYM($$divU_3)
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.export $$divU_3,millicode
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addi 1,x2,x2 /* this CAN overflow */
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addc 0,0,x1
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shd x1,x2,30,t1 /* multiply by 5 to get started */
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sh2add x2,x2,x2
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b LREF(pos)
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addc x1,t1,x1
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/* DIVISION BY 5 (use z = 2**32; a = 33333333) */
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GSYM($$divI_5)
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.export $$divI_5,millicode
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comb,<,N x2,0,LREF(neg5)
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addi 3,x2,t1 /* this cannot overflow */
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sh1add x2,t1,x2 /* multiply by 3 to get started */
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b LREF(pos)
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addc 0,0,x1
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LSYM(neg5)
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sub 0,x2,x2 /* negate x2 */
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addi 1,x2,x2 /* this cannot overflow */
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shd 0,x2,31,x1 /* get top bit (can be 1) */
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sh1add x2,x2,x2 /* multiply by 3 to get started */
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b LREF(neg)
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addc x1,0,x1
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GSYM($$divU_5)
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.export $$divU_5,millicode
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addi 1,x2,x2 /* this CAN overflow */
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addc 0,0,x1
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shd x1,x2,31,t1 /* multiply by 3 to get started */
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sh1add x2,x2,x2
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b LREF(pos)
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addc t1,x1,x1
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/* DIVISION BY 6 (shift to divide by 2 then divide by 3) */
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GSYM($$divI_6)
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.export $$divI_6,millicode
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comb,<,N x2,0,LREF(neg6)
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extru x2,30,31,x2 /* divide by 2 */
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addi 5,x2,t1 /* compute 5*(x2+1) = 5*x2+5 */
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sh2add x2,t1,x2 /* multiply by 5 to get started */
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b LREF(pos)
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addc 0,0,x1
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LSYM(neg6)
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subi 2,x2,x2 /* negate, divide by 2, and add 1 */
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/* negation and adding 1 are done */
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/* at the same time by the SUBI */
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extru x2,30,31,x2
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shd 0,x2,30,x1
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sh2add x2,x2,x2 /* multiply by 5 to get started */
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b LREF(neg)
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addc x1,0,x1
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GSYM($$divU_6)
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.export $$divU_6,millicode
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extru x2,30,31,x2 /* divide by 2 */
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addi 1,x2,x2 /* cannot carry */
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shd 0,x2,30,x1 /* multiply by 5 to get started */
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sh2add x2,x2,x2
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b LREF(pos)
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addc x1,0,x1
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/* DIVISION BY 10 (shift to divide by 2 then divide by 5) */
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GSYM($$divU_10)
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.export $$divU_10,millicode
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extru x2,30,31,x2 /* divide by 2 */
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addi 3,x2,t1 /* compute 3*(x2+1) = (3*x2)+3 */
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sh1add x2,t1,x2 /* multiply by 3 to get started */
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addc 0,0,x1
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LSYM(pos)
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shd x1,x2,28,t1 /* multiply by 0x11 */
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shd x2,0,28,t2
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add x2,t2,x2
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addc x1,t1,x1
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LSYM(pos_for_17)
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shd x1,x2,24,t1 /* multiply by 0x101 */
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shd x2,0,24,t2
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add x2,t2,x2
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addc x1,t1,x1
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shd x1,x2,16,t1 /* multiply by 0x10001 */
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shd x2,0,16,t2
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add x2,t2,x2
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MILLIRET
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addc x1,t1,x1
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GSYM($$divI_10)
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.export $$divI_10,millicode
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comb,< x2,0,LREF(neg10)
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copy 0,x1
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extru x2,30,31,x2 /* divide by 2 */
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addib,TR 1,x2,LREF(pos) /* add 1 (cannot overflow) */
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sh1add x2,x2,x2 /* multiply by 3 to get started */
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LSYM(neg10)
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subi 2,x2,x2 /* negate, divide by 2, and add 1 */
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/* negation and adding 1 are done */
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/* at the same time by the SUBI */
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extru x2,30,31,x2
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sh1add x2,x2,x2 /* multiply by 3 to get started */
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LSYM(neg)
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shd x1,x2,28,t1 /* multiply by 0x11 */
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shd x2,0,28,t2
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add x2,t2,x2
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addc x1,t1,x1
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LSYM(neg_for_17)
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shd x1,x2,24,t1 /* multiply by 0x101 */
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shd x2,0,24,t2
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add x2,t2,x2
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addc x1,t1,x1
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shd x1,x2,16,t1 /* multiply by 0x10001 */
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shd x2,0,16,t2
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add x2,t2,x2
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addc x1,t1,x1
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MILLIRET
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sub 0,x1,x1
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/* DIVISION BY 12 (shift to divide by 4 then divide by 3) */
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GSYM($$divI_12)
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.export $$divI_12,millicode
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comb,< x2,0,LREF(neg12)
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copy 0,x1
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extru x2,29,30,x2 /* divide by 4 */
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addib,tr 1,x2,LREF(pos) /* compute 5*(x2+1) = 5*x2+5 */
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sh2add x2,x2,x2 /* multiply by 5 to get started */
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LSYM(neg12)
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subi 4,x2,x2 /* negate, divide by 4, and add 1 */
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/* negation and adding 1 are done */
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/* at the same time by the SUBI */
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extru x2,29,30,x2
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b LREF(neg)
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sh2add x2,x2,x2 /* multiply by 5 to get started */
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GSYM($$divU_12)
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.export $$divU_12,millicode
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extru x2,29,30,x2 /* divide by 4 */
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addi 5,x2,t1 /* cannot carry */
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sh2add x2,t1,x2 /* multiply by 5 to get started */
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b LREF(pos)
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addc 0,0,x1
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/* DIVISION BY 15 (use z = 2**32; a = 11111111) */
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GSYM($$divI_15)
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.export $$divI_15,millicode
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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
|