Add the rt linux 4.1.3-rt3 as base

Import the rt linux 4.1.3-rt3 as OPNFV kvm base.

It's from git://git.kernel.org/pub/scm/linux/kernel/git/rt/linux-rt-devel.git linux-4.1.y-rt and
the base is:

commit 0917f823c59692d751951bf5ea699a2d1e2f26a2
Author: Sebastian Andrzej Siewior <bigeasy@linutronix.de>
Date:   Sat Jul 25 12:13:34 2015 +0200

    Prepare v4.1.3-rt3

    Signed-off-by: Sebastian Andrzej Siewior <bigeasy@linutronix.de>

We lose all the git history this way and it's not good. We
should apply another opnfv project repo in future.

Change-Id: I87543d81c9df70d99c5001fbdf646b202c19f423
Signed-off-by: Yunhong Jiang <yunhong.jiang@intel.com>

1 files changed, 277 insertions, 0 deletions

diff --git a/kernel/arch/blackfin/lib/udivsi3.S b/kernel/arch/blackfin/lib/udivsi3.S
new file mode 100644
index 000000000..748a6a2e8
--- /dev/null
+++ b/kernel/arch/blackfin/lib/udivsi3.S
@@ -0,0 +1,277 @@
+/*
+ * 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)