root/arch/alpha/lib/ev6-copy_user.S

   1 /* SPDX-License-Identifier: GPL-2.0 */
   2 /*
   3  * arch/alpha/lib/ev6-copy_user.S
   4  *
   5  * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
   6  *
   7  * Copy to/from user space, handling exceptions as we go..  This
   8  * isn't exactly pretty.
   9  *
  10  * This is essentially the same as "memcpy()", but with a few twists.
  11  * Notably, we have to make sure that $0 is always up-to-date and
  12  * contains the right "bytes left to copy" value (and that it is updated
  13  * only _after_ a successful copy). There is also some rather minor
  14  * exception setup stuff..
  15  *
  16  * Much of the information about 21264 scheduling/coding comes from:
  17  *      Compiler Writer's Guide for the Alpha 21264
  18  *      abbreviated as 'CWG' in other comments here
  19  *      ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
  20  * Scheduling notation:
  21  *      E       - either cluster
  22  *      U       - upper subcluster; U0 - subcluster U0; U1 - subcluster U1
  23  *      L       - lower subcluster; L0 - subcluster L0; L1 - subcluster L1
  24  */
  25 
  26 #include <asm/export.h>
  27 /* Allow an exception for an insn; exit if we get one.  */
  28 #define EXI(x,y...)                     \
  29         99: x,##y;                      \
  30         .section __ex_table,"a";        \
  31         .long 99b - .;                  \
  32         lda $31, $exitin-99b($31);      \
  33         .previous
  34 
  35 #define EXO(x,y...)                     \
  36         99: x,##y;                      \
  37         .section __ex_table,"a";        \
  38         .long 99b - .;                  \
  39         lda $31, $exitout-99b($31);     \
  40         .previous
  41 
  42         .set noat
  43         .align 4
  44         .globl __copy_user
  45         .ent __copy_user
  46                                 # Pipeline info: Slotting & Comments
  47 __copy_user:
  48         .prologue 0
  49         mov $18, $0             # .. .. .. E
  50         subq $18, 32, $1        # .. .. E. ..   : Is this going to be a small copy?
  51         nop                     # .. E  .. ..
  52         beq $18, $zerolength    # U  .. .. ..   : U L U L
  53 
  54         and $16,7,$3            # .. .. .. E    : is leading dest misalignment
  55         ble $1, $onebyteloop    # .. .. U  ..   : 1st branch : small amount of data
  56         beq $3, $destaligned    # .. U  .. ..   : 2nd (one cycle fetcher stall)
  57         subq $3, 8, $3          # E  .. .. ..   : L U U L : trip counter
  58 /*
  59  * The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U)
  60  * This loop aligns the destination a byte at a time
  61  * We know we have at least one trip through this loop
  62  */
  63 $aligndest:
  64         EXI( ldbu $1,0($17) )   # .. .. .. L    : Keep loads separate from stores
  65         addq $16,1,$16          # .. .. E  ..   : Section 3.8 in the CWG
  66         addq $3,1,$3            # .. E  .. ..   :
  67         nop                     # E  .. .. ..   : U L U L
  68 
  69 /*
  70  * the -1 is to compensate for the inc($16) done in a previous quadpack
  71  * which allows us zero dependencies within either quadpack in the loop
  72  */
  73         EXO( stb $1,-1($16) )   # .. .. .. L    :
  74         addq $17,1,$17          # .. .. E  ..   : Section 3.8 in the CWG
  75         subq $0,1,$0            # .. E  .. ..   :
  76         bne $3, $aligndest      # U  .. .. ..   : U L U L
  77 
  78 /*
  79  * If we fell through into here, we have a minimum of 33 - 7 bytes
  80  * If we arrived via branch, we have a minimum of 32 bytes
  81  */
  82 $destaligned:
  83         and $17,7,$1            # .. .. .. E    : Check _current_ source alignment
  84         bic $0,7,$4             # .. .. E  ..   : number bytes as a quadword loop
  85         EXI( ldq_u $3,0($17) )  # .. L  .. ..   : Forward fetch for fallthrough code
  86         beq $1,$quadaligned     # U  .. .. ..   : U L U L
  87 
  88 /*
  89  * In the worst case, we've just executed an ldq_u here from 0($17)
  90  * and we'll repeat it once if we take the branch
  91  */
  92 
  93 /* Misaligned quadword loop - not unrolled.  Leave it that way. */
  94 $misquad:
  95         EXI( ldq_u $2,8($17) )  # .. .. .. L    :
  96         subq $4,8,$4            # .. .. E  ..   :
  97         extql $3,$17,$3         # .. U  .. ..   :
  98         extqh $2,$17,$1         # U  .. .. ..   : U U L L
  99 
 100         bis $3,$1,$1            # .. .. .. E    :
 101         EXO( stq $1,0($16) )    # .. .. L  ..   :
 102         addq $17,8,$17          # .. E  .. ..   :
 103         subq $0,8,$0            # E  .. .. ..   : U L L U
 104 
 105         addq $16,8,$16          # .. .. .. E    :
 106         bis $2,$2,$3            # .. .. E  ..   :
 107         nop                     # .. E  .. ..   :
 108         bne $4,$misquad         # U  .. .. ..   : U L U L
 109 
 110         nop                     # .. .. .. E
 111         nop                     # .. .. E  ..
 112         nop                     # .. E  .. ..
 113         beq $0,$zerolength      # U  .. .. ..   : U L U L
 114 
 115 /* We know we have at least one trip through the byte loop */
 116         EXI ( ldbu $2,0($17) )  # .. .. .. L    : No loads in the same quad
 117         addq $16,1,$16          # .. .. E  ..   : as the store (Section 3.8 in CWG)
 118         nop                     # .. E  .. ..   :
 119         br $31, $dirtyentry     # L0 .. .. ..   : L U U L
 120 /* Do the trailing byte loop load, then hop into the store part of the loop */
 121 
 122 /*
 123  * A minimum of (33 - 7) bytes to do a quad at a time.
 124  * Based upon the usage context, it's worth the effort to unroll this loop
 125  * $0 - number of bytes to be moved
 126  * $4 - number of bytes to move as quadwords
 127  * $16 is current destination address
 128  * $17 is current source address
 129  */
 130 $quadaligned:
 131         subq    $4, 32, $2      # .. .. .. E    : do not unroll for small stuff
 132         nop                     # .. .. E  ..
 133         nop                     # .. E  .. ..
 134         blt     $2, $onequad    # U  .. .. ..   : U L U L
 135 
 136 /*
 137  * There is a significant assumption here that the source and destination
 138  * addresses differ by more than 32 bytes.  In this particular case, a
 139  * sparsity of registers further bounds this to be a minimum of 8 bytes.
 140  * But if this isn't met, then the output result will be incorrect.
 141  * Furthermore, due to a lack of available registers, we really can't
 142  * unroll this to be an 8x loop (which would enable us to use the wh64
 143  * instruction memory hint instruction).
 144  */
 145 $unroll4:
 146         EXI( ldq $1,0($17) )    # .. .. .. L
 147         EXI( ldq $2,8($17) )    # .. .. L  ..
 148         subq    $4,32,$4        # .. E  .. ..
 149         nop                     # E  .. .. ..   : U U L L
 150 
 151         addq    $17,16,$17      # .. .. .. E
 152         EXO( stq $1,0($16) )    # .. .. L  ..
 153         EXO( stq $2,8($16) )    # .. L  .. ..
 154         subq    $0,16,$0        # E  .. .. ..   : U L L U
 155 
 156         addq    $16,16,$16      # .. .. .. E
 157         EXI( ldq $1,0($17) )    # .. .. L  ..
 158         EXI( ldq $2,8($17) )    # .. L  .. ..
 159         subq    $4, 32, $3      # E  .. .. ..   : U U L L : is there enough for another trip?
 160 
 161         EXO( stq $1,0($16) )    # .. .. .. L
 162         EXO( stq $2,8($16) )    # .. .. L  ..
 163         subq    $0,16,$0        # .. E  .. ..
 164         addq    $17,16,$17      # E  .. .. ..   : U L L U
 165 
 166         nop                     # .. .. .. E
 167         nop                     # .. .. E  ..
 168         addq    $16,16,$16      # .. E  .. ..
 169         bgt     $3,$unroll4     # U  .. .. ..   : U L U L
 170 
 171         nop
 172         nop
 173         nop
 174         beq     $4, $noquads
 175 
 176 $onequad:
 177         EXI( ldq $1,0($17) )
 178         subq    $4,8,$4
 179         addq    $17,8,$17
 180         nop
 181 
 182         EXO( stq $1,0($16) )
 183         subq    $0,8,$0
 184         addq    $16,8,$16
 185         bne     $4,$onequad
 186 
 187 $noquads:
 188         nop
 189         nop
 190         nop
 191         beq $0,$zerolength
 192 
 193 /*
 194  * For small copies (or the tail of a larger copy), do a very simple byte loop.
 195  * There's no point in doing a lot of complex alignment calculations to try to
 196  * to quadword stuff for a small amount of data.
 197  *      $0 - remaining number of bytes left to copy
 198  *      $16 - current dest addr
 199  *      $17 - current source addr
 200  */
 201 
 202 $onebyteloop:
 203         EXI ( ldbu $2,0($17) )  # .. .. .. L    : No loads in the same quad
 204         addq $16,1,$16          # .. .. E  ..   : as the store (Section 3.8 in CWG)
 205         nop                     # .. E  .. ..   :
 206         nop                     # E  .. .. ..   : U L U L
 207 
 208 $dirtyentry:
 209 /*
 210  * the -1 is to compensate for the inc($16) done in a previous quadpack
 211  * which allows us zero dependencies within either quadpack in the loop
 212  */
 213         EXO ( stb $2,-1($16) )  # .. .. .. L    :
 214         addq $17,1,$17          # .. .. E  ..   : quadpack as the load
 215         subq $0,1,$0            # .. E  .. ..   : change count _after_ copy
 216         bgt $0,$onebyteloop     # U  .. .. ..   : U L U L
 217 
 218 $zerolength:
 219 $exitin:
 220 $exitout:                       # Destination for exception recovery(?)
 221         nop                     # .. .. .. E
 222         nop                     # .. .. E  ..
 223         nop                     # .. E  .. ..
 224         ret $31,($26),1         # L0 .. .. ..   : L U L U
 225 
 226         .end __copy_user
 227         EXPORT_SYMBOL(__copy_user)