/*
* 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
#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;
ENDPROC(___udivsi3)