From b0a68dc07ec395d44849ce98eb417713ca333410 Mon Sep 17 00:00:00 2001
From: Mike Frysinger
Date: Sun, 21 Oct 2007 22:57:36 +0800
Subject: Blackfin arch: add assembly function for doing 64bit unsigned
division
Signed-off-by: Mike Frysinger
Signed-off-by: Bryan Wu
---
arch/blackfin/lib/Makefile | 2 +-
arch/blackfin/lib/udivdi3.S | 375 ++++++++++++++++++++++++++++++++++++++++++++
2 files changed, 376 insertions(+), 1 deletion(-)
create mode 100644 arch/blackfin/lib/udivdi3.S
(limited to 'arch')
diff --git a/arch/blackfin/lib/Makefile b/arch/blackfin/lib/Makefile
index 635288fc5f54..bfdad52c570b 100644
--- a/arch/blackfin/lib/Makefile
+++ b/arch/blackfin/lib/Makefile
@@ -4,7 +4,7 @@
lib-y := \
ashldi3.o ashrdi3.o lshrdi3.o \
- muldi3.o divsi3.o udivsi3.o modsi3.o umodsi3.o \
+ muldi3.o divsi3.o udivsi3.o udivdi3.o modsi3.o umodsi3.o \
checksum.o memcpy.o memset.o memcmp.o memchr.o memmove.o \
strcmp.o strcpy.o strncmp.o strncpy.o \
umulsi3_highpart.o smulsi3_highpart.o \
diff --git a/arch/blackfin/lib/udivdi3.S b/arch/blackfin/lib/udivdi3.S
new file mode 100644
index 000000000000..ad1ebee675e1
--- /dev/null
+++ b/arch/blackfin/lib/udivdi3.S
@@ -0,0 +1,375 @@
+/*
+ * udivdi3.S - unsigned long long division
+ *
+ * Copyright 2003-2007 Analog Devices Inc.
+ * Enter bugs at http://blackfin.uclinux.org/
+ *
+ * Licensed under the GPLv2 or later.
+ */
+
+#include
+
+#define CARRY AC0
+
+#ifdef CONFIG_ARITHMETIC_OPS_L1
+.section .l1.text
+#else
+.text
+#endif
+
+
+ENTRY(___udivdi3)
+ R3 = [SP + 12];
+ [--SP] = (R7:4, P5:3);
+
+ /* Attempt to use divide primitive first; these will handle
+ ** most cases, and they're quick - avoids stalls incurred by
+ ** testing for identities.
+ */
+
+ R4 = R2 | R3;
+ CC = R4 == 0;
+ IF CC JUMP .LDIV_BY_ZERO;
+
+ R4.H = 0x8000;
+ R4 >>>= 16; // R4 now 0xFFFF8000
+ R5 = R0 | R2; // If either dividend or
+ R4 = R5 & R4; // divisor have bits in
+ CC = R4; // top half or low half's sign
+ IF CC JUMP .LIDENTS; // bit, skip builtins.
+ R4 = R1 | R3; // Also check top halves
+ CC = R4;
+ IF CC JUMP .LIDENTS;
+
+ /* Can use the builtins. */
+
+ AQ = CC; // Clear AQ (CC==0)
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ DIVQ(R0, R2);
+ R0 = R0.L (Z);
+ R1 = 0;
+ (R7:4, P5:3) = [SP++];
+ RTS;
+
+.LIDENTS:
+ /* Test for common identities. Value to be returned is
+ ** placed in R6,R7.
+ */
+ // Check for 0/y, return 0
+ R4 = R0 | R1;
+ CC = R4 == 0;
+ IF CC JUMP .LRETURN_R0;
+
+ // Check for x/x, return 1
+ R6 = R0 - R2; // If x == y, then both R6 and R7 will be zero
+ R7 = R1 - R3;
+ R4 = R6 | R7; // making R4 zero.
+ R6 += 1; // which would now make R6:R7==1.
+ CC = R4 == 0;
+ IF CC JUMP .LRETURN_IDENT;
+
+ // Check for x/1, return x
+ R6 = R0;
+ R7 = R1;
+ CC = R3 == 0;
+ IF !CC JUMP .Lnexttest;
+ CC = R2 == 1;
+ IF CC JUMP .LRETURN_IDENT;
+
+.Lnexttest:
+ R4.L = ONES R2; // check for div by power of two which
+ R5.L = ONES R3; // can be done using a shift
+ R6 = PACK (R5.L, R4.L);
+ CC = R6 == 1;
+ IF CC JUMP .Lpower_of_two_upper_zero;
+ R6 = PACK (R4.L, R5.L);
+ CC = R6 == 1;
+ IF CC JUMP .Lpower_of_two_lower_zero;
+
+ // Check for x < y, return 0
+ R6 = 0;
+ R7 = R6;
+ CC = R1 < R3 (IU);
+ IF CC JUMP .LRETURN_IDENT;
+ CC = R1 == R3;
+ IF !CC JUMP .Lno_idents;
+ CC = R0 < R2 (IU);
+ IF CC JUMP .LRETURN_IDENT;
+
+.Lno_idents: // Idents don't match. Go for the full operation
+
+
+ // If X, or X and Y have high bit set, it'll affect the
+ // results, so shift right one to stop this. Note: we've already
+ // checked that X >= Y, so Y's msb won't be set unless X's
+ // is.
+
+ R4 = 0;
+ CC = R1 < 0;
+ IF !CC JUMP .Lx_msb_clear;
+ CC = !CC; // 1 -> 0;
+ R1 = ROT R1 BY -1; // Shift X >> 1
+ R0 = ROT R0 BY -1; // lsb -> CC
+ BITSET(R4,31); // to record only x msb was set
+ CC = R3 < 0;
+ IF !CC JUMP .Ly_msb_clear;
+ CC = !CC;
+ R3 = ROT R3 BY -1; // Shift Y >> 1
+ R2 = ROT R2 BY -1;
+ BITCLR(R4,31); // clear bit to record only x msb was set
+
+.Ly_msb_clear:
+.Lx_msb_clear:
+ // Bit 31 in R4 indicates X msb set, but Y msb wasn't, and no bits
+ // were lost, so we should shift result left by one.
+
+ [--SP] = R4; // save for later
+
+ // In the loop that follows, each iteration we add
+ // either Y' or -Y' to the Remainder. We compute the
+ // negated Y', and store, for convenience. Y' goes
+ // into P0:P1, while -Y' goes into P2:P3.
+
+ P0 = R2;
+ P1 = R3;
+ R2 = -R2;
+ CC = CARRY;
+ CC = !CC;
+ R4 = CC;
+ R3 = -R3;
+ R3 = R3 - R4;
+
+ R6 = 0; // remainder = 0
+ R7 = R6;
+
+ [--SP] = R2; P2 = SP;
+ [--SP] = R3; P3 = SP;
+ [--SP] = R6; P5 = SP; // AQ = 0
+ [--SP] = P1;
+
+ /* In the loop that follows, we use the following
+ ** register assignments:
+ ** R0,R1 X, workspace
+ ** R2,R3 Y, workspace
+ ** R4,R5 partial Div
+ ** R6,R7 partial remainder
+ ** P5 AQ
+ ** The remainder and div form a 128-bit number, with
+ ** the remainder in the high 64-bits.
+ */
+ R4 = R0; // Div = X'
+ R5 = R1;
+ R3 = 0;
+
+ P4 = 64; // Iterate once per bit
+ LSETUP(.LULST,.LULEND) LC0 = P4;
+.LULST:
+ /* Shift Div and remainder up by one. The bit shifted
+ ** out of the top of the quotient is shifted into the bottom
+ ** of the remainder.
+ */
+ CC = R3;
+ R4 = ROT R4 BY 1;
+ R5 = ROT R5 BY 1 || // low q to high q
+ R2 = [P5]; // load saved AQ
+ R6 = ROT R6 BY 1 || // high q to low r
+ R0 = [P2]; // load -Y'
+ R7 = ROT R7 BY 1 || // low r to high r
+ R1 = [P3];
+
+ // Assume add -Y'
+ CC = R2 < 0; // But if AQ is set...
+ IF CC R0 = P0; // then add Y' instead
+ IF CC R1 = P1;
+
+ R6 = R6 + R0; // Rem += (Y' or -Y')
+ CC = CARRY;
+ R0 = CC;
+ R7 = R7 + R1;
+ R7 = R7 + R0 (NS) ||
+ R1 = [SP];
+ // Set the next AQ bit
+ R1 = R7 ^ R1; // from Remainder and Y'
+ R1 = R1 >> 31 || // Negate AQ's value, and
+ [P5] = R1; // save next AQ
+ BITTGL(R1, 0); // add neg AQ to the Div
+.LULEND: R4 = R4 + R1;
+
+ R6 = [SP + 16];
+
+ R0 = R4;
+ R1 = R5;
+ CC = BITTST(R6,30); // Just set CC=0
+ R4 = ROT R0 BY 1; // but if we had to shift X,
+ R5 = ROT R1 BY 1; // and didn't shift any bits out,
+ CC = BITTST(R6,31); // then the result will be half as
+ IF CC R0 = R4; // much as required, so shift left
+ IF CC R1 = R5; // one space.
+
+ SP += 20;
+ (R7:4, P5:3) = [SP++];
+ 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
+ */
+.Lpower_of_two_lower_zero:
+ R7 = 0;
+ R6 = R1 >> 31;
+ CC = R3 < 0;
+ IF CC JUMP .LRETURN_IDENT;
+
+ R2.L = SIGNBITS R3;
+ R2 = R2.L (Z);
+ R2 += -62;
+ (R7:4, P5:3) = [SP++];
+ JUMP ___lshftli;
+
+.Lpower_of_two_upper_zero:
+ CC = R2 < 0;
+ IF CC JUMP .Lmaxint_shift;
+
+ R2.L = SIGNBITS R2;
+ R2 = R2.L (Z);
+ R2 += -30;
+ (R7:4, P5:3) = [SP++];
+ JUMP ___lshftli;
+
+.Lmaxint_shift:
+ R2 = -31;
+ (R7:4, P5:3) = [SP++];
+ JUMP ___lshftli;
+
+.LRETURN_IDENT:
+ R0 = R6;
+ R1 = R7;
+.LRETURN_R0:
+ (R7:4, P5:3) = [SP++];
+ RTS;
+.LDIV_BY_ZERO:
+ R0 = ~R2;
+ R1 = R0;
+ (R7:4, P5:3) = [SP++];
+ RTS;
+
+ENDPROC(___udivdi3)
+
+
+ENTRY(___lshftli)
+ CC = R2 == 0;
+ IF CC JUMP .Lfinished; // nothing to do
+ CC = R2 < 0;
+ IF CC JUMP .Lrshift;
+ R3 = 64;
+ CC = R2 < R3;
+ IF !CC JUMP .Lretzero;
+
+ // We're shifting left, and it's less than 64 bits, so
+ // a valid result will be returned.
+
+ R3 >>= 1; // R3 now 32
+ CC = R2 < R3;
+
+ IF !CC JUMP .Lzerohalf;
+
+ // We're shifting left, between 1 and 31 bits, which means
+ // some of the low half will be shifted into the high half.
+ // Work out how much.
+
+ R3 = R3 - R2;
+
+ // Save that much data from the bottom half.
+
+ P1 = R7;
+ R7 = R0;
+ R7 >>= R3;
+
+ // Adjust both parts of the parameter.
+
+ R0 <<= R2;
+ R1 <<= R2;
+
+ // And include the bits moved across.
+
+ R1 = R1 | R7;
+ R7 = P1;
+ RTS;
+
+.Lzerohalf:
+ // We're shifting left, between 32 and 63 bits, so the
+ // bottom half will become zero, and the top half will
+ // lose some bits. How many?
+
+ R2 = R2 - R3; // N - 32
+ R1 = LSHIFT R0 BY R2.L;
+ R0 = R0 - R0;
+ RTS;
+
+.Lretzero:
+ R0 = R0 - R0;
+ R1 = R0;
+.Lfinished:
+ RTS;
+
+.Lrshift:
+ // We're shifting right, but by how much?
+ R2 = -R2;
+ R3 = 64;
+ CC = R2 < R3;
+ IF !CC JUMP .Lretzero;
+
+ // Shifting right less than 64 bits, so some result bits will
+ // be retained.
+
+ R3 >>= 1; // R3 now 32
+ CC = R2 < R3;
+ IF !CC JUMP .Lsignhalf;
+
+ // Shifting right between 1 and 31 bits, so need to copy
+ // data across words.
+
+ P1 = R7;
+ R3 = R3 - R2;
+ R7 = R1;
+ R7 <<= R3;
+ R1 >>= R2;
+ R0 >>= R2;
+ R0 = R7 | R0;
+ R7 = P1;
+ RTS;
+
+.Lsignhalf:
+ // Shifting right between 32 and 63 bits, so the top half
+ // will become all zero-bits, and the bottom half is some
+ // of the top half. But how much?
+
+ R2 = R2 - R3;
+ R0 = R1;
+ R0 >>= R2;
+ R1 = 0;
+ RTS;
+
+ENDPROC(___lshftli)
--
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