kernel-ark/arch/blackfin/lib/udivsi3.S
Bryan Wu 1394f03221 blackfin architecture
This adds support for the Analog Devices Blackfin processor architecture, and
currently supports the BF533, BF532, BF531, BF537, BF536, BF534, and BF561
(Dual Core) devices, with a variety of development platforms including those
avaliable from Analog Devices (BF533-EZKit, BF533-STAMP, BF537-STAMP,
BF561-EZKIT), and Bluetechnix!  Tinyboards.

The Blackfin architecture was jointly developed by Intel and Analog Devices
Inc.  (ADI) as the Micro Signal Architecture (MSA) core and introduced it in
December of 2000.  Since then ADI has put this core into its Blackfin
processor family of devices.  The Blackfin core has the advantages of a clean,
orthogonal,RISC-like microprocessor instruction set.  It combines a dual-MAC
(Multiply/Accumulate), state-of-the-art signal processing engine and
single-instruction, multiple-data (SIMD) multimedia capabilities into a single
instruction-set architecture.

The Blackfin architecture, including the instruction set, is described by the
ADSP-BF53x/BF56x Blackfin Processor Programming Reference
http://blackfin.uclinux.org/gf/download/frsrelease/29/2549/Blackfin_PRM.pdf

The Blackfin processor is already supported by major releases of gcc, and
there are binary and source rpms/tarballs for many architectures at:
http://blackfin.uclinux.org/gf/project/toolchain/frs There is complete
documentation, including "getting started" guides available at:
http://docs.blackfin.uclinux.org/ which provides links to the sources and
patches you will need in order to set up a cross-compiling environment for
bfin-linux-uclibc

This patch, as well as the other patches (toolchain, distribution,
uClibc) are actively supported by Analog Devices Inc, at:
http://blackfin.uclinux.org/

We have tested this on LTP, and our test plan (including pass/fails) can
be found at:
http://docs.blackfin.uclinux.org/doku.php?id=testing_the_linux_kernel

[m.kozlowski@tuxland.pl: balance parenthesis in blackfin header files]
Signed-off-by: Bryan Wu <bryan.wu@analog.com>
Signed-off-by: Mariusz Kozlowski <m.kozlowski@tuxland.pl>
Signed-off-by: Aubrey Li <aubrey.li@analog.com>
Signed-off-by: Jie Zhang <jie.zhang@analog.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
2007-05-07 12:12:58 -07:00

299 lines
9.1 KiB
ArmAsm

/*
* File: arch/blackfin/lib/udivsi3.S
* Based on:
* Author:
*
* Created:
* Description:
*
* Modified:
* Copyright 2004-2006 Analog Devices Inc.
*
* Bugs: Enter bugs at http://blackfin.uclinux.org/
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; either version 2 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, see the file COPYING, or write
* to the Free Software Foundation, Inc.,
* 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
*/
#include <linux/linkage.h>
#define CARRY AC0
#ifdef CONFIG_ARITHMETIC_OPS_L1
.section .l1.text
#else
.text
#endif
ENTRY(___udivsi3)
CC = R0 < R1 (IU); /* If X < Y, always return 0 */
IF CC JUMP .Lreturn_ident;
R2 = R1 << 16;
CC = R2 <= R0 (IU);
IF CC JUMP .Lidents;
R2 = R0 >> 31; /* if X is a 31-bit number */
R3 = R1 >> 15; /* and Y is a 15-bit number */
R2 = R2 | R3; /* then it's okay to use the DIVQ builtins (fallthrough to fast)*/
CC = R2;
IF CC JUMP .Ly_16bit;
/* METHOD 1: FAST DIVQ
We know we have a 31-bit dividend, and 15-bit divisor so we can use the
simple divq approach (first setting AQ to 0 - implying unsigned division,
then 16 DIVQ's).
*/
AQ = CC; /* Clear AQ (CC==0) */
/* ISR States: When dividing two integers (32.0/16.0) using divide primitives,
we need to shift the dividend one bit to the left.
We have already checked that we have a 31-bit number so we are safe to do
that.
*/
R0 <<= 1;
DIVQ(R0, R1); // 1
DIVQ(R0, R1); // 2
DIVQ(R0, R1); // 3
DIVQ(R0, R1); // 4
DIVQ(R0, R1); // 5
DIVQ(R0, R1); // 6
DIVQ(R0, R1); // 7
DIVQ(R0, R1); // 8
DIVQ(R0, R1); // 9
DIVQ(R0, R1); // 10
DIVQ(R0, R1); // 11
DIVQ(R0, R1); // 12
DIVQ(R0, R1); // 13
DIVQ(R0, R1); // 14
DIVQ(R0, R1); // 15
DIVQ(R0, R1); // 16
R0 = R0.L (Z);
RTS;
.Ly_16bit:
/* We know that the upper 17 bits of Y might have bits set,
** or that the sign bit of X might have a bit. If Y is a
** 16-bit number, but not bigger, then we can use the builtins
** with a post-divide correction.
** R3 currently holds Y>>15, which means R3's LSB is the
** bit we're interested in.
*/
/* According to the ISR, to use the Divide primitives for
** unsigned integer divide, the useable range is 31 bits
*/
CC = ! BITTST(R0, 31);
/* IF condition is true we can scale our inputs and use the divide primitives,
** with some post-adjustment
*/
R3 += -1; /* if so, Y is 0x00008nnn */
CC &= AZ;
/* If condition is true we can scale our inputs and use the divide primitives,
** with some post-adjustment
*/
R3 = R1 >> 1; /* Pre-scaled divisor for primitive case */
R2 = R0 >> 16;
R2 = R3 - R2; /* shifted divisor < upper 16 bits of dividend */
CC &= CARRY;
IF CC JUMP .Lshift_and_correct;
/* Fall through to the identities */
/* METHOD 2: identities and manual calculation
We are not able to use the divide primites, but may still catch some special
cases.
*/
.Lidents:
/* Test for common identities. Value to be returned is placed in R2. */
CC = R0 == 0; /* 0/Y => 0 */
IF CC JUMP .Lreturn_r0;
CC = R0 == R1; /* X==Y => 1 */
IF CC JUMP .Lreturn_ident;
CC = R1 == 1; /* X/1 => X */
IF CC JUMP .Lreturn_ident;
R2.L = ONES R1;
R2 = R2.L (Z);
CC = R2 == 1;
IF CC JUMP .Lpower_of_two;
[--SP] = (R7:5); /* Push registers R5-R7 */
/* Idents don't match. Go for the full operation. */
R6 = 2; /* assume we'll shift two */
R3 = 1;
P2 = R1;
/* If either R0 or R1 have sign set, */
/* divide them by two, and note it's */
/* been done. */
CC = R1 < 0;
R2 = R1 >> 1;
IF CC R1 = R2; /* Possibly-shifted R1 */
IF !CC R6 = R3; /* R1 doesn't, so at most 1 shifted */
P0 = 0;
R3 = -R1;
[--SP] = R3;
R2 = R0 >> 1;
R2 = R0 >> 1;
CC = R0 < 0;
IF CC P0 = R6; /* Number of values divided */
IF !CC R2 = R0; /* Shifted R0 */
/* P0 is 0, 1 (NR/=2) or 2 (NR/=2, DR/=2) */
/* r2 holds Copy dividend */
R3 = 0; /* Clear partial remainder */
R7 = 0; /* Initialise quotient bit */
P1 = 32; /* Set loop counter */
LSETUP(.Lulst, .Lulend) LC0 = P1; /* Set loop counter */
.Lulst: R6 = R2 >> 31; /* R6 = sign bit of R2, for carry */
R2 = R2 << 1; /* Shift 64 bit dividend up by 1 bit */
R3 = R3 << 1 || R5 = [SP];
R3 = R3 | R6; /* Include any carry */
CC = R7 < 0; /* Check quotient(AQ) */
/* If AQ==0, we'll sub divisor */
IF CC R5 = R1; /* and if AQ==1, we'll add it. */
R3 = R3 + R5; /* Add/sub divsor to partial remainder */
R7 = R3 ^ R1; /* Generate next quotient bit */
R5 = R7 >> 31; /* Get AQ */
BITTGL(R5, 0); /* Invert it, to get what we'll shift */
.Lulend: R2 = R2 + R5; /* and "shift" it in. */
CC = P0 == 0; /* Check how many inputs we shifted */
IF CC JUMP .Lno_mult; /* if none... */
R6 = R2 << 1;
CC = P0 == 1;
IF CC R2 = R6; /* if 1, Q = Q*2 */
IF !CC R1 = P2; /* if 2, restore stored divisor */
R3 = R2; /* Copy of R2 */
R3 *= R1; /* Q * divisor */
R5 = R0 - R3; /* Z = (dividend - Q * divisor) */
CC = R1 <= R5 (IU); /* Check if divisor <= Z? */
R6 = CC; /* if yes, R6 = 1 */
R2 = R2 + R6; /* if yes, add one to quotient(Q) */
.Lno_mult:
SP += 4;
(R7:5) = [SP++]; /* Pop registers R5-R7 */
R0 = R2; /* Store quotient */
RTS;
.Lreturn_ident:
CC = R0 < R1 (IU); /* If X < Y, always return 0 */
R2 = 0;
IF CC JUMP .Ltrue_return_ident;
R2 = -1 (X); /* X/0 => 0xFFFFFFFF */
CC = R1 == 0;
IF CC JUMP .Ltrue_return_ident;
R2 = -R2; /* R2 now 1 */
CC = R0 == R1; /* X==Y => 1 */
IF CC JUMP .Ltrue_return_ident;
R2 = R0; /* X/1 => X */
/*FALLTHRU*/
.Ltrue_return_ident:
R0 = R2;
.Lreturn_r0:
RTS;
.Lpower_of_two:
/* Y has a single bit set, which means it's a power of two.
** That means we can perform the division just by shifting
** X to the right the appropriate number of bits
*/
/* signbits returns the number of sign bits, minus one.
** 1=>30, 2=>29, ..., 0x40000000=>0. Which means we need
** to shift right n-signbits spaces. It also means 0x80000000
** is a special case, because that *also* gives a signbits of 0
*/
R2 = R0 >> 31;
CC = R1 < 0;
IF CC JUMP .Ltrue_return_ident;
R1.l = SIGNBITS R1;
R1 = R1.L (Z);
R1 += -30;
R0 = LSHIFT R0 by R1.L;
RTS;
/* METHOD 3: PRESCALE AND USE THE DIVIDE PRIMITIVES WITH SOME POST-CORRECTION
Two scaling operations are required to use the divide primitives with a
divisor > 0x7FFFF.
Firstly (as in method 1) we need to shift the dividend 1 to the left for
integer division.
Secondly we need to shift both the divisor and dividend 1 to the right so
both are in range for the primitives.
The left/right shift of the dividend does nothing so we can skip it.
*/
.Lshift_and_correct:
R2 = R0;
// R3 is already R1 >> 1
CC=!CC;
AQ = CC; /* Clear AQ, got here with CC = 0 */
DIVQ(R2, R3); // 1
DIVQ(R2, R3); // 2
DIVQ(R2, R3); // 3
DIVQ(R2, R3); // 4
DIVQ(R2, R3); // 5
DIVQ(R2, R3); // 6
DIVQ(R2, R3); // 7
DIVQ(R2, R3); // 8
DIVQ(R2, R3); // 9
DIVQ(R2, R3); // 10
DIVQ(R2, R3); // 11
DIVQ(R2, R3); // 12
DIVQ(R2, R3); // 13
DIVQ(R2, R3); // 14
DIVQ(R2, R3); // 15
DIVQ(R2, R3); // 16
/* According to the Instruction Set Reference:
To divide by a divisor > 0x7FFF,
1. prescale and perform divide to obtain quotient (Q) (done above),
2. multiply quotient by unscaled divisor (result M)
3. subtract the product from the divident to get an error (E = X - M)
4. if E < divisor (Y) subtract 1, if E > divisor (Y) add 1, else return quotient (Q)
*/
R3 = R2.L (Z); /* Q = X' / Y' */
R2 = R3; /* Preserve Q */
R2 *= R1; /* M = Q * Y */
R2 = R0 - R2; /* E = X - M */
R0 = R3; /* Copy Q into result reg */
/* Correction: If result of the multiply is negative, we overflowed
and need to correct the result by subtracting 1 from the result.*/
R3 = 0xFFFF (Z);
R2 = R2 >> 16; /* E >> 16 */
CC = R2 == R3;
R3 = 1 ;
R1 = R0 - R3;
IF CC R0 = R1;
RTS;