kernel / pub / scm / linux / kernel / git / pkl / squashfs-next / ef0a59924a795ccb4ced0ae1722a337745a1b045 / . / arch / blackfin / lib / udivsi3.S

/* | |

* Copyright 2004-2009 Analog Devices Inc. | |

* | |

* Licensed under the Clear 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) |