Donate to e Foundation | Murena handsets with /e/OS | Own a part of Murena! Learn more

Commit 1754a5d9 authored by Mike Frysinger's avatar Mike Frysinger Committed by Bryan Wu
Browse files

Blackfin arch: use do_div() for the 64bit division as pointed out by Bernd 



If you need a 64 bit divide in the kernel, use asm/div64.h.
Revert the addition of udivdi3.

Cc: Bernd Schmidt <bernd.schmidt@analog.com>
Signed-off-by: default avatarMike Frysinger <michael.frysinger@analog.com>
Signed-off-by: default avatarBryan Wu <bryan.wu@analog.com>
parent e709d84b

arch/blackfin/kernel/setup.c

+7 −2
Original line number Diff line number Diff line
@@ -43,6 +43,7 @@ 
#include <asm/cacheflush.h>

#include <asm/blackfin.h>

#include <asm/cplbinit.h>

#include <asm/div64.h>

#include <asm/fixed_code.h>

#include <asm/early_printk.h>
@@ -504,13 +505,17 @@ EXPORT_SYMBOL(get_sclk); 


unsigned long sclk_to_usecs(unsigned long sclk)

{

	return (USEC_PER_SEC * (u64)sclk) / get_sclk();

	u64 tmp = USEC_PER_SEC * (u64)sclk;

	do_div(tmp, get_sclk());

	return tmp;

}

EXPORT_SYMBOL(sclk_to_usecs);



unsigned long usecs_to_sclk(unsigned long usecs)

{

	return (get_sclk() * (u64)usecs) / USEC_PER_SEC;

	u64 tmp = get_sclk() * (u64)usecs;

	do_div(tmp, USEC_PER_SEC);

	return tmp;

}

EXPORT_SYMBOL(usecs_to_sclk);

arch/blackfin/lib/Makefile

+1 −1
Original line number Diff line number Diff line
@@ -4,7 +4,7 @@ 


lib-y := \

	ashldi3.o ashrdi3.o lshrdi3.o \

	muldi3.o divsi3.o udivsi3.o udivdi3.o modsi3.o umodsi3.o \

	muldi3.o divsi3.o udivsi3.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 \

arch/blackfin/lib/udivdi3.S

deleted100644 → 0
+0 −375
Original line number Diff line number Diff line 
/*

 * 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 <linux/linkage.h>



#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)