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https://github.com/AetherDroid/android_kernel_samsung_on5xelte.git
synced 2025-09-09 01:28:05 -04:00
Fixed MTP to work with TWRP
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awab228
commit
f6dfaef42e
50820 changed files with 20846062 additions and 0 deletions
277
arch/blackfin/lib/udivsi3.S
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277
arch/blackfin/lib/udivsi3.S
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/*
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* Copyright 2004-2009 Analog Devices Inc.
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*
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* Licensed under the Clear BSD license or the GPL-2 (or later)
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*/
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#include <linux/linkage.h>
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#define CARRY AC0
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#ifdef CONFIG_ARITHMETIC_OPS_L1
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.section .l1.text
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#else
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.text
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#endif
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ENTRY(___udivsi3)
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CC = R0 < R1 (IU); /* If X < Y, always return 0 */
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IF CC JUMP .Lreturn_ident;
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R2 = R1 << 16;
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CC = R2 <= R0 (IU);
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IF CC JUMP .Lidents;
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R2 = R0 >> 31; /* if X is a 31-bit number */
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R3 = R1 >> 15; /* and Y is a 15-bit number */
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R2 = R2 | R3; /* then it's okay to use the DIVQ builtins (fallthrough to fast)*/
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CC = R2;
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IF CC JUMP .Ly_16bit;
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/* METHOD 1: FAST DIVQ
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We know we have a 31-bit dividend, and 15-bit divisor so we can use the
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simple divq approach (first setting AQ to 0 - implying unsigned division,
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then 16 DIVQ's).
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*/
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AQ = CC; /* Clear AQ (CC==0) */
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/* ISR States: When dividing two integers (32.0/16.0) using divide primitives,
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we need to shift the dividend one bit to the left.
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We have already checked that we have a 31-bit number so we are safe to do
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that.
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*/
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R0 <<= 1;
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DIVQ(R0, R1); // 1
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DIVQ(R0, R1); // 2
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DIVQ(R0, R1); // 3
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DIVQ(R0, R1); // 4
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DIVQ(R0, R1); // 5
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DIVQ(R0, R1); // 6
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DIVQ(R0, R1); // 7
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DIVQ(R0, R1); // 8
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DIVQ(R0, R1); // 9
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DIVQ(R0, R1); // 10
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DIVQ(R0, R1); // 11
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DIVQ(R0, R1); // 12
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DIVQ(R0, R1); // 13
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DIVQ(R0, R1); // 14
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DIVQ(R0, R1); // 15
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DIVQ(R0, R1); // 16
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R0 = R0.L (Z);
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RTS;
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.Ly_16bit:
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/* We know that the upper 17 bits of Y might have bits set,
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** or that the sign bit of X might have a bit. If Y is a
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** 16-bit number, but not bigger, then we can use the builtins
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** with a post-divide correction.
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** R3 currently holds Y>>15, which means R3's LSB is the
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** bit we're interested in.
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*/
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/* According to the ISR, to use the Divide primitives for
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** unsigned integer divide, the useable range is 31 bits
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*/
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CC = ! BITTST(R0, 31);
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/* IF condition is true we can scale our inputs and use the divide primitives,
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** with some post-adjustment
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*/
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R3 += -1; /* if so, Y is 0x00008nnn */
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CC &= AZ;
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/* If condition is true we can scale our inputs and use the divide primitives,
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** with some post-adjustment
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*/
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R3 = R1 >> 1; /* Pre-scaled divisor for primitive case */
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R2 = R0 >> 16;
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R2 = R3 - R2; /* shifted divisor < upper 16 bits of dividend */
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CC &= CARRY;
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IF CC JUMP .Lshift_and_correct;
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/* Fall through to the identities */
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/* METHOD 2: identities and manual calculation
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We are not able to use the divide primites, but may still catch some special
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cases.
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*/
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.Lidents:
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/* Test for common identities. Value to be returned is placed in R2. */
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CC = R0 == 0; /* 0/Y => 0 */
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IF CC JUMP .Lreturn_r0;
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CC = R0 == R1; /* X==Y => 1 */
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IF CC JUMP .Lreturn_ident;
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CC = R1 == 1; /* X/1 => X */
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IF CC JUMP .Lreturn_ident;
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R2.L = ONES R1;
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R2 = R2.L (Z);
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CC = R2 == 1;
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IF CC JUMP .Lpower_of_two;
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[--SP] = (R7:5); /* Push registers R5-R7 */
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/* Idents don't match. Go for the full operation. */
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R6 = 2; /* assume we'll shift two */
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R3 = 1;
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P2 = R1;
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/* If either R0 or R1 have sign set, */
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/* divide them by two, and note it's */
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/* been done. */
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CC = R1 < 0;
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R2 = R1 >> 1;
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IF CC R1 = R2; /* Possibly-shifted R1 */
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IF !CC R6 = R3; /* R1 doesn't, so at most 1 shifted */
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P0 = 0;
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R3 = -R1;
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[--SP] = R3;
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R2 = R0 >> 1;
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R2 = R0 >> 1;
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CC = R0 < 0;
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IF CC P0 = R6; /* Number of values divided */
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IF !CC R2 = R0; /* Shifted R0 */
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/* P0 is 0, 1 (NR/=2) or 2 (NR/=2, DR/=2) */
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/* r2 holds Copy dividend */
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R3 = 0; /* Clear partial remainder */
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R7 = 0; /* Initialise quotient bit */
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P1 = 32; /* Set loop counter */
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LSETUP(.Lulst, .Lulend) LC0 = P1; /* Set loop counter */
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.Lulst: R6 = R2 >> 31; /* R6 = sign bit of R2, for carry */
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R2 = R2 << 1; /* Shift 64 bit dividend up by 1 bit */
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R3 = R3 << 1 || R5 = [SP];
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R3 = R3 | R6; /* Include any carry */
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CC = R7 < 0; /* Check quotient(AQ) */
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/* If AQ==0, we'll sub divisor */
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IF CC R5 = R1; /* and if AQ==1, we'll add it. */
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R3 = R3 + R5; /* Add/sub divsor to partial remainder */
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R7 = R3 ^ R1; /* Generate next quotient bit */
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R5 = R7 >> 31; /* Get AQ */
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BITTGL(R5, 0); /* Invert it, to get what we'll shift */
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.Lulend: R2 = R2 + R5; /* and "shift" it in. */
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CC = P0 == 0; /* Check how many inputs we shifted */
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IF CC JUMP .Lno_mult; /* if none... */
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R6 = R2 << 1;
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CC = P0 == 1;
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IF CC R2 = R6; /* if 1, Q = Q*2 */
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IF !CC R1 = P2; /* if 2, restore stored divisor */
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R3 = R2; /* Copy of R2 */
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R3 *= R1; /* Q * divisor */
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R5 = R0 - R3; /* Z = (dividend - Q * divisor) */
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CC = R1 <= R5 (IU); /* Check if divisor <= Z? */
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R6 = CC; /* if yes, R6 = 1 */
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R2 = R2 + R6; /* if yes, add one to quotient(Q) */
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.Lno_mult:
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SP += 4;
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(R7:5) = [SP++]; /* Pop registers R5-R7 */
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R0 = R2; /* Store quotient */
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RTS;
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.Lreturn_ident:
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CC = R0 < R1 (IU); /* If X < Y, always return 0 */
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R2 = 0;
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IF CC JUMP .Ltrue_return_ident;
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R2 = -1 (X); /* X/0 => 0xFFFFFFFF */
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CC = R1 == 0;
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IF CC JUMP .Ltrue_return_ident;
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R2 = -R2; /* R2 now 1 */
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CC = R0 == R1; /* X==Y => 1 */
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IF CC JUMP .Ltrue_return_ident;
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R2 = R0; /* X/1 => X */
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/*FALLTHRU*/
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.Ltrue_return_ident:
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R0 = R2;
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.Lreturn_r0:
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RTS;
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.Lpower_of_two:
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/* Y has a single bit set, which means it's a power of two.
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** That means we can perform the division just by shifting
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** X to the right the appropriate number of bits
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*/
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/* signbits returns the number of sign bits, minus one.
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** 1=>30, 2=>29, ..., 0x40000000=>0. Which means we need
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** to shift right n-signbits spaces. It also means 0x80000000
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** is a special case, because that *also* gives a signbits of 0
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*/
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R2 = R0 >> 31;
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CC = R1 < 0;
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IF CC JUMP .Ltrue_return_ident;
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R1.l = SIGNBITS R1;
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R1 = R1.L (Z);
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R1 += -30;
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R0 = LSHIFT R0 by R1.L;
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RTS;
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/* METHOD 3: PRESCALE AND USE THE DIVIDE PRIMITIVES WITH SOME POST-CORRECTION
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Two scaling operations are required to use the divide primitives with a
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divisor > 0x7FFFF.
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Firstly (as in method 1) we need to shift the dividend 1 to the left for
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integer division.
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Secondly we need to shift both the divisor and dividend 1 to the right so
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both are in range for the primitives.
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The left/right shift of the dividend does nothing so we can skip it.
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*/
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.Lshift_and_correct:
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R2 = R0;
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// R3 is already R1 >> 1
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CC=!CC;
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AQ = CC; /* Clear AQ, got here with CC = 0 */
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DIVQ(R2, R3); // 1
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DIVQ(R2, R3); // 2
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DIVQ(R2, R3); // 3
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DIVQ(R2, R3); // 4
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DIVQ(R2, R3); // 5
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DIVQ(R2, R3); // 6
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DIVQ(R2, R3); // 7
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DIVQ(R2, R3); // 8
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DIVQ(R2, R3); // 9
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DIVQ(R2, R3); // 10
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DIVQ(R2, R3); // 11
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DIVQ(R2, R3); // 12
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DIVQ(R2, R3); // 13
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DIVQ(R2, R3); // 14
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DIVQ(R2, R3); // 15
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DIVQ(R2, R3); // 16
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/* According to the Instruction Set Reference:
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To divide by a divisor > 0x7FFF,
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1. prescale and perform divide to obtain quotient (Q) (done above),
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2. multiply quotient by unscaled divisor (result M)
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3. subtract the product from the divident to get an error (E = X - M)
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4. if E < divisor (Y) subtract 1, if E > divisor (Y) add 1, else return quotient (Q)
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*/
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R3 = R2.L (Z); /* Q = X' / Y' */
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R2 = R3; /* Preserve Q */
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R2 *= R1; /* M = Q * Y */
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R2 = R0 - R2; /* E = X - M */
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R0 = R3; /* Copy Q into result reg */
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/* Correction: If result of the multiply is negative, we overflowed
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and need to correct the result by subtracting 1 from the result.*/
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R3 = 0xFFFF (Z);
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R2 = R2 >> 16; /* E >> 16 */
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CC = R2 == R3;
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R3 = 1 ;
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R1 = R0 - R3;
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IF CC R0 = R1;
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RTS;
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ENDPROC(___udivsi3)
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