udivsi3.s

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/* * 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#endifENTRY(___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)

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