Adding upstream version 6.1.76.upstream/6.1.76upstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

4 files changed, 933 insertions, 0 deletions

diff --git a/Documentation/driver-api/mtd/index.rst b/Documentation/driver-api/mtd/index.rst
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+++ b/Documentation/driver-api/mtd/index.rst
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+.. SPDX-License-Identifier: GPL-2.0
+
+==============================
+Memory Technology Device (MTD)
+==============================
+
+.. toctree::
+   :maxdepth: 1
+
+   spi-intel
+   nand_ecc
+   spi-nor
diff --git a/Documentation/driver-api/mtd/nand_ecc.rst b/Documentation/driver-api/mtd/nand_ecc.rst
new file mode 100644
index 000000000..74347c14a
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+==========================
+NAND Error-correction Code
+==========================
+
+Introduction
+============
+
+Having looked at the linux mtd/nand Hamming software ECC engine driver
+I felt there was room for optimisation. I bashed the code for a few hours
+performing tricks like table lookup removing superfluous code etc.
+After that the speed was increased by 35-40%.
+Still I was not too happy as I felt there was additional room for improvement.
+
+Bad! I was hooked.
+I decided to annotate my steps in this file. Perhaps it is useful to someone
+or someone learns something from it.
+
+
+The problem
+===========
+
+NAND flash (at least SLC one) typically has sectors of 256 bytes.
+However NAND flash is not extremely reliable so some error detection
+(and sometimes correction) is needed.
+
+This is done by means of a Hamming code. I'll try to explain it in
+laymans terms (and apologies to all the pro's in the field in case I do
+not use the right terminology, my coding theory class was almost 30
+years ago, and I must admit it was not one of my favourites).
+
+As I said before the ecc calculation is performed on sectors of 256
+bytes. This is done by calculating several parity bits over the rows and
+columns. The parity used is even parity which means that the parity bit = 1
+if the data over which the parity is calculated is 1 and the parity bit = 0
+if the data over which the parity is calculated is 0. So the total
+number of bits over the data over which the parity is calculated + the
+parity bit is even. (see wikipedia if you can't follow this).
+Parity is often calculated by means of an exclusive or operation,
+sometimes also referred to as xor. In C the operator for xor is ^
+
+Back to ecc.
+Let's give a small figure:
+
+=========  ==== ==== ==== ==== ==== ==== ==== ====   === === === === ====
+byte   0:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp0 rp2 rp4 ... rp14
+byte   1:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp1 rp2 rp4 ... rp14
+byte   2:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp0 rp3 rp4 ... rp14
+byte   3:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp1 rp3 rp4 ... rp14
+byte   4:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp0 rp2 rp5 ... rp14
+...
+byte 254:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp0 rp3 rp5 ... rp15
+byte 255:  bit7 bit6 bit5 bit4 bit3 bit2 bit1 bit0   rp1 rp3 rp5 ... rp15
+           cp1  cp0  cp1  cp0  cp1  cp0  cp1  cp0
+           cp3  cp3  cp2  cp2  cp3  cp3  cp2  cp2
+           cp5  cp5  cp5  cp5  cp4  cp4  cp4  cp4
+=========  ==== ==== ==== ==== ==== ==== ==== ====   === === === === ====
+
+This figure represents a sector of 256 bytes.
+cp is my abbreviation for column parity, rp for row parity.
+
+Let's start to explain column parity.
+
+- cp0 is the parity that belongs to all bit0, bit2, bit4, bit6.
+
+  so the sum of all bit0, bit2, bit4 and bit6 values + cp0 itself is even.
+
+Similarly cp1 is the sum of all bit1, bit3, bit5 and bit7.
+
+- cp2 is the parity over bit0, bit1, bit4 and bit5
+- cp3 is the parity over bit2, bit3, bit6 and bit7.
+- cp4 is the parity over bit0, bit1, bit2 and bit3.
+- cp5 is the parity over bit4, bit5, bit6 and bit7.
+
+Note that each of cp0 .. cp5 is exactly one bit.
+
+Row parity actually works almost the same.
+
+- rp0 is the parity of all even bytes (0, 2, 4, 6, ... 252, 254)
+- rp1 is the parity of all odd bytes (1, 3, 5, 7, ..., 253, 255)
+- rp2 is the parity of all bytes 0, 1, 4, 5, 8, 9, ...
+  (so handle two bytes, then skip 2 bytes).
+- rp3 is covers the half rp2 does not cover (bytes 2, 3, 6, 7, 10, 11, ...)
+- for rp4 the rule is cover 4 bytes, skip 4 bytes, cover 4 bytes, skip 4 etc.
+
+  so rp4 calculates parity over bytes 0, 1, 2, 3, 8, 9, 10, 11, 16, ...)
+- and rp5 covers the other half, so bytes 4, 5, 6, 7, 12, 13, 14, 15, 20, ..
+
+The story now becomes quite boring. I guess you get the idea.
+
+- rp6 covers 8 bytes then skips 8 etc
+- rp7 skips 8 bytes then covers 8 etc
+- rp8 covers 16 bytes then skips 16 etc
+- rp9 skips 16 bytes then covers 16 etc
+- rp10 covers 32 bytes then skips 32 etc
+- rp11 skips 32 bytes then covers 32 etc
+- rp12 covers 64 bytes then skips 64 etc
+- rp13 skips 64 bytes then covers 64 etc
+- rp14 covers 128 bytes then skips 128
+- rp15 skips 128 bytes then covers 128
+
+In the end the parity bits are grouped together in three bytes as
+follows:
+
+=====  ===== ===== ===== ===== ===== ===== ===== =====
+ECC    Bit 7 Bit 6 Bit 5 Bit 4 Bit 3 Bit 2 Bit 1 Bit 0
+=====  ===== ===== ===== ===== ===== ===== ===== =====
+ECC 0   rp07  rp06  rp05  rp04  rp03  rp02  rp01  rp00
+ECC 1   rp15  rp14  rp13  rp12  rp11  rp10  rp09  rp08
+ECC 2   cp5   cp4   cp3   cp2   cp1   cp0      1     1
+=====  ===== ===== ===== ===== ===== ===== ===== =====
+
+I detected after writing this that ST application note AN1823
+(http://www.st.com/stonline/) gives a much
+nicer picture.(but they use line parity as term where I use row parity)
+Oh well, I'm graphically challenged, so suffer with me for a moment :-)
+
+And I could not reuse the ST picture anyway for copyright reasons.
+
+
+Attempt 0
+=========
+
+Implementing the parity calculation is pretty simple.
+In C pseudocode::
+
+  for (i = 0; i < 256; i++)
+  {
+    if (i & 0x01)
+       rp1 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp1;
+    else
+       rp0 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp0;
+    if (i & 0x02)
+       rp3 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp3;
+    else
+       rp2 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp2;
+    if (i & 0x04)
+      rp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp5;
+    else
+      rp4 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp4;
+    if (i & 0x08)
+      rp7 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp7;
+    else
+      rp6 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp6;
+    if (i & 0x10)
+      rp9 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp9;
+    else
+      rp8 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp8;
+    if (i & 0x20)
+      rp11 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp11;
+    else
+      rp10 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp10;
+    if (i & 0x40)
+      rp13 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp13;
+    else
+      rp12 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp12;
+    if (i & 0x80)
+      rp15 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp15;
+    else
+      rp14 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ bit3 ^ bit2 ^ bit1 ^ bit0 ^ rp14;
+    cp0 = bit6 ^ bit4 ^ bit2 ^ bit0 ^ cp0;
+    cp1 = bit7 ^ bit5 ^ bit3 ^ bit1 ^ cp1;
+    cp2 = bit5 ^ bit4 ^ bit1 ^ bit0 ^ cp2;
+    cp3 = bit7 ^ bit6 ^ bit3 ^ bit2 ^ cp3
+    cp4 = bit3 ^ bit2 ^ bit1 ^ bit0 ^ cp4
+    cp5 = bit7 ^ bit6 ^ bit5 ^ bit4 ^ cp5
+  }
+
+
+Analysis 0
+==========
+
+C does have bitwise operators but not really operators to do the above
+efficiently (and most hardware has no such instructions either).
+Therefore without implementing this it was clear that the code above was
+not going to bring me a Nobel prize :-)
+
+Fortunately the exclusive or operation is commutative, so we can combine
+the values in any order. So instead of calculating all the bits
+individually, let us try to rearrange things.
+For the column parity this is easy. We can just xor the bytes and in the
+end filter out the relevant bits. This is pretty nice as it will bring
+all cp calculation out of the for loop.
+
+Similarly we can first xor the bytes for the various rows.
+This leads to:
+
+
+Attempt 1
+=========
+
+::
+
+  const char parity[256] = {
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1,
+      0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
+  };
+
+  void ecc1(const unsigned char *buf, unsigned char *code)
+  {
+      int i;
+      const unsigned char *bp = buf;
+      unsigned char cur;
+      unsigned char rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
+      unsigned char rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
+      unsigned char par;
+
+      par = 0;
+      rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
+      rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
+      rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
+      rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
+
+      for (i = 0; i < 256; i++)
+      {
+          cur = *bp++;
+          par ^= cur;
+          if (i & 0x01) rp1 ^= cur; else rp0 ^= cur;
+          if (i & 0x02) rp3 ^= cur; else rp2 ^= cur;
+          if (i & 0x04) rp5 ^= cur; else rp4 ^= cur;
+          if (i & 0x08) rp7 ^= cur; else rp6 ^= cur;
+          if (i & 0x10) rp9 ^= cur; else rp8 ^= cur;
+          if (i & 0x20) rp11 ^= cur; else rp10 ^= cur;
+          if (i & 0x40) rp13 ^= cur; else rp12 ^= cur;
+          if (i & 0x80) rp15 ^= cur; else rp14 ^= cur;
+      }
+      code[0] =
+          (parity[rp7] << 7) |
+          (parity[rp6] << 6) |
+          (parity[rp5] << 5) |
+          (parity[rp4] << 4) |
+          (parity[rp3] << 3) |
+          (parity[rp2] << 2) |
+          (parity[rp1] << 1) |
+          (parity[rp0]);
+      code[1] =
+          (parity[rp15] << 7) |
+          (parity[rp14] << 6) |
+          (parity[rp13] << 5) |
+          (parity[rp12] << 4) |
+          (parity[rp11] << 3) |
+          (parity[rp10] << 2) |
+          (parity[rp9]  << 1) |
+          (parity[rp8]);
+      code[2] =
+          (parity[par & 0xf0] << 7) |
+          (parity[par & 0x0f] << 6) |
+          (parity[par & 0xcc] << 5) |
+          (parity[par & 0x33] << 4) |
+          (parity[par & 0xaa] << 3) |
+          (parity[par & 0x55] << 2);
+      code[0] = ~code[0];
+      code[1] = ~code[1];
+      code[2] = ~code[2];
+  }
+
+Still pretty straightforward. The last three invert statements are there to
+give a checksum of 0xff 0xff 0xff for an empty flash. In an empty flash
+all data is 0xff, so the checksum then matches.
+
+I also introduced the parity lookup. I expected this to be the fastest
+way to calculate the parity, but I will investigate alternatives later
+on.
+
+
+Analysis 1
+==========
+
+The code works, but is not terribly efficient. On my system it took
+almost 4 times as much time as the linux driver code. But hey, if it was
+*that* easy this would have been done long before.
+No pain. no gain.
+
+Fortunately there is plenty of room for improvement.
+
+In step 1 we moved from bit-wise calculation to byte-wise calculation.
+However in C we can also use the unsigned long data type and virtually
+every modern microprocessor supports 32 bit operations, so why not try
+to write our code in such a way that we process data in 32 bit chunks.
+
+Of course this means some modification as the row parity is byte by
+byte. A quick analysis:
+for the column parity we use the par variable. When extending to 32 bits
+we can in the end easily calculate rp0 and rp1 from it.
+(because par now consists of 4 bytes, contributing to rp1, rp0, rp1, rp0
+respectively, from MSB to LSB)
+also rp2 and rp3 can be easily retrieved from par as rp3 covers the
+first two MSBs and rp2 covers the last two LSBs.
+
+Note that of course now the loop is executed only 64 times (256/4).
+And note that care must taken wrt byte ordering. The way bytes are
+ordered in a long is machine dependent, and might affect us.
+Anyway, if there is an issue: this code is developed on x86 (to be
+precise: a DELL PC with a D920 Intel CPU)
+
+And of course the performance might depend on alignment, but I expect
+that the I/O buffers in the nand driver are aligned properly (and
+otherwise that should be fixed to get maximum performance).
+
+Let's give it a try...
+
+
+Attempt 2
+=========
+
+::
+
+  extern const char parity[256];
+
+  void ecc2(const unsigned char *buf, unsigned char *code)
+  {
+      int i;
+      const unsigned long *bp = (unsigned long *)buf;
+      unsigned long cur;
+      unsigned long rp0, rp1, rp2, rp3, rp4, rp5, rp6, rp7;
+      unsigned long rp8, rp9, rp10, rp11, rp12, rp13, rp14, rp15;
+      unsigned long par;
+
+      par = 0;
+      rp0 = 0; rp1 = 0; rp2 = 0; rp3 = 0;
+      rp4 = 0; rp5 = 0; rp6 = 0; rp7 = 0;
+      rp8 = 0; rp9 = 0; rp10 = 0; rp11 = 0;
+      rp12 = 0; rp13 = 0; rp14 = 0; rp15 = 0;
+
+      for (i = 0; i < 64; i++)
+      {
+          cur = *bp++;
+          par ^= cur;
+          if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
+          if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
+          if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
+          if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
+          if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
+          if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
+      }
+      /*
+         we need to adapt the code generation for the fact that rp vars are now
+         long; also the column parity calculation needs to be changed.
+         we'll bring rp4 to 15 back to single byte entities by shifting and
+         xoring
+      */
+      rp4 ^= (rp4 >> 16); rp4 ^= (rp4 >> 8); rp4 &= 0xff;
+      rp5 ^= (rp5 >> 16); rp5 ^= (rp5 >> 8); rp5 &= 0xff;
+      rp6 ^= (rp6 >> 16); rp6 ^= (rp6 >> 8); rp6 &= 0xff;
+      rp7 ^= (rp7 >> 16); rp7 ^= (rp7 >> 8); rp7 &= 0xff;
+      rp8 ^= (rp8 >> 16); rp8 ^= (rp8 >> 8); rp8 &= 0xff;
+      rp9 ^= (rp9 >> 16); rp9 ^= (rp9 >> 8); rp9 &= 0xff;
+      rp10 ^= (rp10 >> 16); rp10 ^= (rp10 >> 8); rp10 &= 0xff;
+      rp11 ^= (rp11 >> 16); rp11 ^= (rp11 >> 8); rp11 &= 0xff;
+      rp12 ^= (rp12 >> 16); rp12 ^= (rp12 >> 8); rp12 &= 0xff;
+      rp13 ^= (rp13 >> 16); rp13 ^= (rp13 >> 8); rp13 &= 0xff;
+      rp14 ^= (rp14 >> 16); rp14 ^= (rp14 >> 8); rp14 &= 0xff;
+      rp15 ^= (rp15 >> 16); rp15 ^= (rp15 >> 8); rp15 &= 0xff;
+      rp3 = (par >> 16); rp3 ^= (rp3 >> 8); rp3 &= 0xff;
+      rp2 = par & 0xffff; rp2 ^= (rp2 >> 8); rp2 &= 0xff;
+      par ^= (par >> 16);
+      rp1 = (par >> 8); rp1 &= 0xff;
+      rp0 = (par & 0xff);
+      par ^= (par >> 8); par &= 0xff;
+
+      code[0] =
+          (parity[rp7] << 7) |
+          (parity[rp6] << 6) |
+          (parity[rp5] << 5) |
+          (parity[rp4] << 4) |
+          (parity[rp3] << 3) |
+          (parity[rp2] << 2) |
+          (parity[rp1] << 1) |
+          (parity[rp0]);
+      code[1] =
+          (parity[rp15] << 7) |
+          (parity[rp14] << 6) |
+          (parity[rp13] << 5) |
+          (parity[rp12] << 4) |
+          (parity[rp11] << 3) |
+          (parity[rp10] << 2) |
+          (parity[rp9]  << 1) |
+          (parity[rp8]);
+      code[2] =
+          (parity[par & 0xf0] << 7) |
+          (parity[par & 0x0f] << 6) |
+          (parity[par & 0xcc] << 5) |
+          (parity[par & 0x33] << 4) |
+          (parity[par & 0xaa] << 3) |
+          (parity[par & 0x55] << 2);
+      code[0] = ~code[0];
+      code[1] = ~code[1];
+      code[2] = ~code[2];
+  }
+
+The parity array is not shown any more. Note also that for these
+examples I kinda deviated from my regular programming style by allowing
+multiple statements on a line, not using { } in then and else blocks
+with only a single statement and by using operators like ^=
+
+
+Analysis 2
+==========
+
+The code (of course) works, and hurray: we are a little bit faster than
+the linux driver code (about 15%). But wait, don't cheer too quickly.
+There is more to be gained.
+If we look at e.g. rp14 and rp15 we see that we either xor our data with
+rp14 or with rp15. However we also have par which goes over all data.
+This means there is no need to calculate rp14 as it can be calculated from
+rp15 through rp14 = par ^ rp15, because par = rp14 ^ rp15;
+(or if desired we can avoid calculating rp15 and calculate it from
+rp14).  That is why some places refer to inverse parity.
+Of course the same thing holds for rp4/5, rp6/7, rp8/9, rp10/11 and rp12/13.
+Effectively this means we can eliminate the else clause from the if
+statements. Also we can optimise the calculation in the end a little bit
+by going from long to byte first. Actually we can even avoid the table
+lookups
+
+Attempt 3
+=========
+
+Odd replaced::
+
+          if (i & 0x01) rp5 ^= cur; else rp4 ^= cur;
+          if (i & 0x02) rp7 ^= cur; else rp6 ^= cur;
+          if (i & 0x04) rp9 ^= cur; else rp8 ^= cur;
+          if (i & 0x08) rp11 ^= cur; else rp10 ^= cur;
+          if (i & 0x10) rp13 ^= cur; else rp12 ^= cur;
+          if (i & 0x20) rp15 ^= cur; else rp14 ^= cur;
+
+with::
+
+          if (i & 0x01) rp5 ^= cur;
+          if (i & 0x02) rp7 ^= cur;
+          if (i & 0x04) rp9 ^= cur;
+          if (i & 0x08) rp11 ^= cur;
+          if (i & 0x10) rp13 ^= cur;
+          if (i & 0x20) rp15 ^= cur;
+
+and outside the loop added::
+
+          rp4  = par ^ rp5;
+          rp6  = par ^ rp7;
+          rp8  = par ^ rp9;
+          rp10  = par ^ rp11;
+          rp12  = par ^ rp13;
+          rp14  = par ^ rp15;
+
+And after that the code takes about 30% more time, although the number of
+statements is reduced. This is also reflected in the assembly code.
+
+
+Analysis 3
+==========
+
+Very weird. Guess it has to do with caching or instruction parallellism
+or so. I also tried on an eeePC (Celeron, clocked at 900 Mhz). Interesting
+observation was that this one is only 30% slower (according to time)
+executing the code as my 3Ghz D920 processor.
+
+Well, it was expected not to be easy so maybe instead move to a
+different track: let's move back to the code from attempt2 and do some
+loop unrolling. This will eliminate a few if statements. I'll try
+different amounts of unrolling to see what works best.
+
+
+Attempt 4
+=========
+
+Unrolled the loop 1, 2, 3 and 4 times.
+For 4 the code starts with::
+
+    for (i = 0; i < 4; i++)
+    {
+        cur = *bp++;
+        par ^= cur;
+        rp4 ^= cur;
+        rp6 ^= cur;
+        rp8 ^= cur;
+        rp10 ^= cur;
+        if (i & 0x1) rp13 ^= cur; else rp12 ^= cur;
+        if (i & 0x2) rp15 ^= cur; else rp14 ^= cur;
+        cur = *bp++;
+        par ^= cur;
+        rp5 ^= cur;
+        rp6 ^= cur;
+        ...
+
+
+Analysis 4
+==========
+
+Unrolling once gains about 15%
+
+Unrolling twice keeps the gain at about 15%
+
+Unrolling three times gives a gain of 30% compared to attempt 2.
+
+Unrolling four times gives a marginal improvement compared to unrolling
+three times.
+
+I decided to proceed with a four time unrolled loop anyway. It was my gut
+feeling that in the next steps I would obtain additional gain from it.
+
+The next step was triggered by the fact that par contains the xor of all
+bytes and rp4 and rp5 each contain the xor of half of the bytes.
+So in effect par = rp4 ^ rp5. But as xor is commutative we can also say
+that rp5 = par ^ rp4. So no need to keep both rp4 and rp5 around. We can
+eliminate rp5 (or rp4, but I already foresaw another optimisation).
+The same holds for rp6/7, rp8/9, rp10/11 rp12/13 and rp14/15.
+
+
+Attempt 5
+=========
+
+Effectively so all odd digit rp assignments in the loop were removed.
+This included the else clause of the if statements.
+Of course after the loop we need to correct things by adding code like::
+
+    rp5 = par ^ rp4;
+
+Also the initial assignments (rp5 = 0; etc) could be removed.
+Along the line I also removed the initialisation of rp0/1/2/3.
+
+
+Analysis 5
+==========
+
+Measurements showed this was a good move. The run-time roughly halved
+compared with attempt 4 with 4 times unrolled, and we only require 1/3rd
+of the processor time compared to the current code in the linux kernel.
+
+However, still I thought there was more. I didn't like all the if
+statements. Why not keep a running parity and only keep the last if
+statement. Time for yet another version!
+
+
+Attempt 6
+=========
+
+THe code within the for loop was changed to::
+
+    for (i = 0; i < 4; i++)
+    {
+        cur = *bp++; tmppar  = cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
+
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
+
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur; rp8 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur; rp8 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp8 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp8 ^= cur;
+
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur;
+
+        par ^= tmppar;
+        if ((i & 0x1) == 0) rp12 ^= tmppar;
+        if ((i & 0x2) == 0) rp14 ^= tmppar;
+    }
+
+As you can see tmppar is used to accumulate the parity within a for
+iteration. In the last 3 statements is added to par and, if needed,
+to rp12 and rp14.
+
+While making the changes I also found that I could exploit that tmppar
+contains the running parity for this iteration. So instead of having:
+rp4 ^= cur; rp6 ^= cur;
+I removed the rp6 ^= cur; statement and did rp6 ^= tmppar; on next
+statement. A similar change was done for rp8 and rp10
+
+
+Analysis 6
+==========
+
+Measuring this code again showed big gain. When executing the original
+linux code 1 million times, this took about 1 second on my system.
+(using time to measure the performance). After this iteration I was back
+to 0.075 sec. Actually I had to decide to start measuring over 10
+million iterations in order not to lose too much accuracy. This one
+definitely seemed to be the jackpot!
+
+There is a little bit more room for improvement though. There are three
+places with statements::
+
+	rp4 ^= cur; rp6 ^= cur;
+
+It seems more efficient to also maintain a variable rp4_6 in the while
+loop; This eliminates 3 statements per loop. Of course after the loop we
+need to correct by adding::
+
+	rp4 ^= rp4_6;
+	rp6 ^= rp4_6
+
+Furthermore there are 4 sequential assignments to rp8. This can be
+encoded slightly more efficiently by saving tmppar before those 4 lines
+and later do rp8 = rp8 ^ tmppar ^ notrp8;
+(where notrp8 is the value of rp8 before those 4 lines).
+Again a use of the commutative property of xor.
+Time for a new test!
+
+
+Attempt 7
+=========
+
+The new code now looks like::
+
+    for (i = 0; i < 4; i++)
+    {
+        cur = *bp++; tmppar  = cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= tmppar;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp8 ^= tmppar;
+
+        cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp10 ^= tmppar;
+
+        notrp8 = tmppar;
+        cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur;
+        rp8 = rp8 ^ tmppar ^ notrp8;
+
+        cur = *bp++; tmppar ^= cur; rp4_6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp6 ^= cur;
+        cur = *bp++; tmppar ^= cur; rp4 ^= cur;
+        cur = *bp++; tmppar ^= cur;
+
+        par ^= tmppar;
+        if ((i & 0x1) == 0) rp12 ^= tmppar;
+        if ((i & 0x2) == 0) rp14 ^= tmppar;
+    }
+    rp4 ^= rp4_6;
+    rp6 ^= rp4_6;
+
+
+Not a big change, but every penny counts :-)
+
+
+Analysis 7
+==========
+
+Actually this made things worse. Not very much, but I don't want to move
+into the wrong direction. Maybe something to investigate later. Could
+have to do with caching again.
+
+Guess that is what there is to win within the loop. Maybe unrolling one
+more time will help. I'll keep the optimisations from 7 for now.
+
+
+Attempt 8
+=========
+
+Unrolled the loop one more time.
+
+
+Analysis 8
+==========
+
+This makes things worse. Let's stick with attempt 6 and continue from there.
+Although it seems that the code within the loop cannot be optimised
+further there is still room to optimize the generation of the ecc codes.
+We can simply calculate the total parity. If this is 0 then rp4 = rp5
+etc. If the parity is 1, then rp4 = !rp5;
+
+But if rp4 = rp5 we do not need rp5 etc. We can just write the even bits
+in the result byte and then do something like::
+
+    code[0] |= (code[0] << 1);
+
+Lets test this.
+
+
+Attempt 9
+=========
+
+Changed the code but again this slightly degrades performance. Tried all
+kind of other things, like having dedicated parity arrays to avoid the
+shift after parity[rp7] << 7; No gain.
+Change the lookup using the parity array by using shift operators (e.g.
+replace parity[rp7] << 7 with::
+
+	rp7 ^= (rp7 << 4);
+	rp7 ^= (rp7 << 2);
+	rp7 ^= (rp7 << 1);
+	rp7 &= 0x80;
+
+No gain.
+
+The only marginal change was inverting the parity bits, so we can remove
+the last three invert statements.
+
+Ah well, pity this does not deliver more. Then again 10 million
+iterations using the linux driver code takes between 13 and 13.5
+seconds, whereas my code now takes about 0.73 seconds for those 10
+million iterations. So basically I've improved the performance by a
+factor 18 on my system. Not that bad. Of course on different hardware
+you will get different results. No warranties!
+
+But of course there is no such thing as a free lunch. The codesize almost
+tripled (from 562 bytes to 1434 bytes). Then again, it is not that much.
+
+
+Correcting errors
+=================
+
+For correcting errors I again used the ST application note as a starter,
+but I also peeked at the existing code.
+
+The algorithm itself is pretty straightforward. Just xor the given and
+the calculated ecc. If all bytes are 0 there is no problem. If 11 bits
+are 1 we have one correctable bit error. If there is 1 bit 1, we have an
+error in the given ecc code.
+
+It proved to be fastest to do some table lookups. Performance gain
+introduced by this is about a factor 2 on my system when a repair had to
+be done, and 1% or so if no repair had to be done.
+
+Code size increased from 330 bytes to 686 bytes for this function.
+(gcc 4.2, -O3)
+
+
+Conclusion
+==========
+
+The gain when calculating the ecc is tremendous. Om my development hardware
+a speedup of a factor of 18 for ecc calculation was achieved. On a test on an
+embedded system with a MIPS core a factor 7 was obtained.
+
+On a test with a Linksys NSLU2 (ARMv5TE processor) the speedup was a factor
+5 (big endian mode, gcc 4.1.2, -O3)
+
+For correction not much gain could be obtained (as bitflips are rare). Then
+again there are also much less cycles spent there.
+
+It seems there is not much more gain possible in this, at least when
+programmed in C. Of course it might be possible to squeeze something more
+out of it with an assembler program, but due to pipeline behaviour etc
+this is very tricky (at least for intel hw).
+
+Author: Frans Meulenbroeks
+
+Copyright (C) 2008 Koninklijke Philips Electronics NV.
diff --git a/Documentation/driver-api/mtd/spi-intel.rst b/Documentation/driver-api/mtd/spi-intel.rst
new file mode 100644
index 000000000..df854f20e
--- /dev/null
+++ b/Documentation/driver-api/mtd/spi-intel.rst
@@ -0,0 +1,90 @@
+==============================
+Upgrading BIOS using spi-intel
+==============================
+
+Many Intel CPUs like Baytrail and Braswell include SPI serial flash host
+controller which is used to hold BIOS and other platform specific data.
+Since contents of the SPI serial flash is crucial for machine to function,
+it is typically protected by different hardware protection mechanisms to
+avoid accidental (or on purpose) overwrite of the content.
+
+Not all manufacturers protect the SPI serial flash, mainly because it
+allows upgrading the BIOS image directly from an OS.
+
+The spi-intel driver makes it possible to read and write the SPI serial
+flash, if certain protection bits are not set and locked. If it finds
+any of them set, the whole MTD device is made read-only to prevent
+partial overwrites. By default the driver exposes SPI serial flash
+contents as read-only but it can be changed from kernel command line,
+passing "spi_intel.writeable=1".
+
+Please keep in mind that overwriting the BIOS image on SPI serial flash
+might render the machine unbootable and requires special equipment like
+Dediprog to revive. You have been warned!
+
+Below are the steps how to upgrade MinnowBoard MAX BIOS directly from
+Linux.
+
+ 1) Download and extract the latest Minnowboard MAX BIOS SPI image
+    [1]. At the time writing this the latest image is v92.
+
+ 2) Install mtd-utils package [2]. We need this in order to erase the SPI
+    serial flash. Distros like Debian and Fedora have this prepackaged with
+    name "mtd-utils".
+
+ 3) Add "spi_intel.writeable=1" to the kernel command line and reboot
+    the board (you can also reload the driver passing "writeable=1" as
+    module parameter to modprobe).
+
+ 4) Once the board is up and running again, find the right MTD partition
+    (it is named as "BIOS")::
+
+	# cat /proc/mtd
+	dev:    size   erasesize  name
+	mtd0: 00800000 00001000 "BIOS"
+
+    So here it will be /dev/mtd0 but it may vary.
+
+ 5) Make backup of the existing image first::
+
+	# dd if=/dev/mtd0ro of=bios.bak
+	16384+0 records in
+	16384+0 records out
+	8388608 bytes (8.4 MB) copied, 10.0269 s, 837 kB/s
+
+ 6) Verify the backup::
+
+	# sha1sum /dev/mtd0ro bios.bak
+	fdbb011920572ca6c991377c4b418a0502668b73  /dev/mtd0ro
+	fdbb011920572ca6c991377c4b418a0502668b73  bios.bak
+
+    The SHA1 sums must match. Otherwise do not continue any further!
+
+ 7) Erase the SPI serial flash. After this step, do not reboot the
+    board! Otherwise it will not start anymore::
+
+	# flash_erase /dev/mtd0 0 0
+	Erasing 4 Kibyte @ 7ff000 -- 100 % complete
+
+ 8) Once completed without errors you can write the new BIOS image::
+
+    # dd if=MNW2MAX1.X64.0092.R01.1605221712.bin of=/dev/mtd0
+
+ 9) Verify that the new content of the SPI serial flash matches the new
+    BIOS image::
+
+	# sha1sum /dev/mtd0ro MNW2MAX1.X64.0092.R01.1605221712.bin
+	9b4df9e4be2057fceec3a5529ec3d950836c87a2  /dev/mtd0ro
+	9b4df9e4be2057fceec3a5529ec3d950836c87a2 MNW2MAX1.X64.0092.R01.1605221712.bin
+
+    The SHA1 sums should match.
+
+ 10) Now you can reboot your board and observe the new BIOS starting up
+     properly.
+
+References
+----------
+
+[1] https://firmware.intel.com/sites/default/files/MinnowBoard%2EMAX_%2EX64%2E92%2ER01%2Ezip
+
+[2] http://www.linux-mtd.infradead.org/
diff --git a/Documentation/driver-api/mtd/spi-nor.rst b/Documentation/driver-api/mtd/spi-nor.rst
new file mode 100644
index 000000000..4a3adca41
--- /dev/null
+++ b/Documentation/driver-api/mtd/spi-nor.rst
@@ -0,0 +1,68 @@
+=================
+SPI NOR framework
+=================
+
+Part I - Why do we need this framework?
+---------------------------------------
+
+SPI bus controllers (drivers/spi/) only deal with streams of bytes; the bus
+controller operates agnostic of the specific device attached. However, some
+controllers (such as Freescale's QuadSPI controller) cannot easily handle
+arbitrary streams of bytes, but rather are designed specifically for SPI NOR.
+
+In particular, Freescale's QuadSPI controller must know the NOR commands to
+find the right LUT sequence. Unfortunately, the SPI subsystem has no notion of
+opcodes, addresses, or data payloads; a SPI controller simply knows to send or
+receive bytes (Tx and Rx). Therefore, we must define a new layering scheme under
+which the controller driver is aware of the opcodes, addressing, and other
+details of the SPI NOR protocol.
+
+Part II - How does the framework work?
+--------------------------------------
+
+This framework just adds a new layer between the MTD and the SPI bus driver.
+With this new layer, the SPI NOR controller driver does not depend on the
+m25p80 code anymore.
+
+Before this framework, the layer is like::
+
+                   MTD
+         ------------------------
+                  m25p80
+         ------------------------
+	       SPI bus driver
+         ------------------------
+	        SPI NOR chip
+
+After this framework, the layer is like::
+
+                   MTD
+         ------------------------
+              SPI NOR framework
+         ------------------------
+                  m25p80
+         ------------------------
+	       SPI bus driver
+         ------------------------
+	       SPI NOR chip
+
+With the SPI NOR controller driver (Freescale QuadSPI), it looks like::
+
+                   MTD
+         ------------------------
+              SPI NOR framework
+         ------------------------
+                fsl-quadSPI
+         ------------------------
+	       SPI NOR chip
+
+Part III - How can drivers use the framework?
+---------------------------------------------
+
+The main API is spi_nor_scan(). Before you call the hook, a driver should
+initialize the necessary fields for spi_nor{}. Please see
+drivers/mtd/spi-nor/spi-nor.c for detail. Please also refer to spi-fsl-qspi.c
+when you want to write a new driver for a SPI NOR controller.
+Another API is spi_nor_restore(), this is used to restore the status of SPI
+flash chip such as addressing mode. Call it whenever detach the driver from
+device or reboot the system.