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https://github.com/edk2-porting/linux-next.git
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96f1050d3d
Bill Gatliff & David Brownell pointed out we were missing some copyrights, and licensing terms in some of the files in ./arch/blackfin, so this fixes things, and cleans them up. It also removes: - verbose GPL text(refer to the top level ./COPYING file) - file names (you are looking at the file) - bug url (it's in the ./MAINTAINERS file) - "or later" on GPL-2, when we did not have that right It also allows some Blackfin-specific assembly files to be under a BSD like license (for people to use them outside of Linux). Signed-off-by: Robin Getz <robin.getz@analog.com> Signed-off-by: Mike Frysinger <vapier@gentoo.org>
200 lines
5.4 KiB
ArmAsm
200 lines
5.4 KiB
ArmAsm
/*
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* Copyright 2004-2009 Analog Devices Inc.
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*
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* Licensed under the ADI BSD license or the GPL-2 (or later)
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*
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* 16 / 32 bit signed division.
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* Special cases :
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* 1) If(numerator == 0)
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* return 0
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* 2) If(denominator ==0)
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* return positive max = 0x7fffffff
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* 3) If(numerator == denominator)
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* return 1
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* 4) If(denominator ==1)
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* return numerator
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* 5) If(denominator == -1)
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* return -numerator
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*
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* Operand : R0 - Numerator (i)
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* R1 - Denominator (i)
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* R0 - Quotient (o)
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* Registers Used : R2-R7,P0-P2
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*
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*/
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.global ___divsi3;
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.type ___divsi3, STT_FUNC;
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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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.align 2;
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___divsi3 :
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R3 = R0 ^ R1;
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R0 = ABS R0;
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CC = V;
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r3 = rot r3 by -1;
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r1 = abs r1; /* now both positive, r3.30 means "negate result",
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** r3.31 means overflow, add one to result
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*/
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cc = r0 < r1;
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if cc jump .Lret_zero;
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r2 = r1 >> 15;
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cc = r2;
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if cc jump .Lidents;
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r2 = r1 << 16;
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cc = r2 <= r0;
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if cc jump .Lidents;
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DIVS(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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DIVQ(R0, R1);
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R0 = R0.L (Z);
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r1 = r3 >> 31; /* add overflow issue back in */
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r0 = r0 + r1;
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r1 = -r0;
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cc = bittst(r3, 30);
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if cc r0 = r1;
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RTS;
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/* Can't use the primitives. Test common identities.
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** If the identity is true, return the value in R2.
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*/
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.Lidents:
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CC = R1 == 0; /* check for divide by zero */
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IF CC JUMP .Lident_return;
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CC = R0 == 0; /* check for division of zero */
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IF CC JUMP .Lzero_return;
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CC = R0 == R1; /* check for identical operands */
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IF CC JUMP .Lident_return;
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CC = R1 == 1; /* check for divide by 1 */
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IF CC JUMP .Lident_return;
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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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/* Identities haven't helped either.
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** Perform the full division process.
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*/
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P1 = 31; /* Set loop counter */
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[--SP] = (R7:5); /* Push registers R5-R7 */
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R2 = -R1;
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[--SP] = R2;
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R2 = R0 << 1; /* R2 lsw of dividend */
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R6 = R0 ^ R1; /* Get sign */
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R5 = R6 >> 31; /* Shift sign to LSB */
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R0 = 0 ; /* Clear msw partial remainder */
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R2 = R2 | R5; /* Shift quotient bit */
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R6 = R0 ^ R1; /* Get new quotient bit */
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LSETUP(.Llst,.Llend) LC0 = P1; /* Setup loop */
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.Llst: R7 = R2 >> 31; /* record copy of carry from R2 */
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R2 = R2 << 1; /* Shift 64 bit dividend up by 1 bit */
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R0 = R0 << 1 || R5 = [SP];
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R0 = R0 | R7; /* and add carry */
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CC = R6 < 0; /* Check quotient(AQ) */
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/* we might be subtracting divisor (AQ==0) */
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IF CC R5 = R1; /* or we might be adding divisor (AQ==1)*/
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R0 = R0 + R5; /* do add or subtract, as indicated by AQ */
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R6 = R0 ^ R1; /* Generate next quotient bit */
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R5 = R6 >> 31;
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/* Assume AQ==1, shift in zero */
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BITTGL(R5,0); /* tweak AQ to be what we want to shift in */
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.Llend: R2 = R2 + R5; /* and then set shifted-in value to
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** tweaked AQ.
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*/
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r1 = r3 >> 31;
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r2 = r2 + r1;
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cc = bittst(r3,30);
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r0 = -r2;
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if !cc r0 = r2;
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SP += 4;
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(R7:5)= [SP++]; /* Pop registers R6-R7 */
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RTS;
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.Lident_return:
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CC = R1 == 0; /* check for divide by zero => 0x7fffffff */
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R2 = -1 (X);
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R2 >>= 1;
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IF CC JUMP .Ltrue_ident_return;
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CC = R0 == R1; /* check for identical operands => 1 */
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R2 = 1 (Z);
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IF CC JUMP .Ltrue_ident_return;
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R2 = R0; /* assume divide by 1 => numerator */
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/*FALLTHRU*/
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.Ltrue_ident_return:
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R0 = R2; /* Return an identity value */
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R2 = -R2;
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CC = bittst(R3,30);
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IF CC R0 = R2;
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.Lzero_return:
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RTS; /* ...including zero */
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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_ident_return;
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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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r1 = r3 >> 31;
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r0 = r0 + r1;
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R2 = -R0; // negate result if necessary
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CC = bittst(R3,30);
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IF CC R0 = R2;
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RTS;
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.Lret_zero:
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R0 = 0;
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RTS;
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.size ___divsi3, .-___divsi3
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