|  | /* | 
|  | * Copyright 2004-2009 Analog Devices Inc. | 
|  | * | 
|  | * Licensed under the ADI BSD license or the GPL-2 (or later) | 
|  | */ | 
|  |  | 
|  | #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; | 
|  |  | 
|  | ENDPROC(___udivsi3) |