Adding upstream version 1:10.11.6.upstream/1%10.11.6upstream

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

1 files changed, 212 insertions, 0 deletions

diff --git a/sql/sql_acl_getsort.ic b/sql/sql_acl_getsort.ic
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+/* Copyright (c) 2019, MariaDB Corporation.
+
+   This program is free software; you can redistribute it and/or modify
+   it under the terms of the GNU General Public License as published by
+   the Free Software Foundation; version 2 of the License.
+
+   This program is distributed in the hope that it will be useful,
+   but WITHOUT ANY WARRANTY; without even the implied warranty of
+   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+   GNU General Public License for more details.
+
+   You should have received a copy of the GNU General Public License
+   along with this program; if not, write to the Free Software
+   Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1335  USA */
+
+#ifndef NO_EMBEDDED_ACCESS_CHECKS
+
+#define magic_bits 30
+/*
+  Returns a number which, if sorted in descending order, magically puts
+  patterns in the order from most specific (e.g. no wildcards) to most generic
+  (e.g. "%"). That is, the larger the number, the more specific the pattern is.
+
+  Takes a template that lists types of following patterns (by the first letter
+  of _h_ostname, _d_bname, _u_sername) and up to four patterns.
+  No more than two can be of 'h' or 'd' type (because one magic value takes
+  magic_bits bits, see below).
+
+  ========================================================================
+
+  Here's how the magic is created:
+
+  Let's look at one iteration of the for() loop. That's one pattern.  With
+  wildcards (usernames aren't interesting).
+
+  By definition a pattern A is "more specific" than pattern B if the set of
+  strings that match the pattern A is smaller than the set of strings that
+  match the pattern B. Strings are taken from the big superset of all valid
+  utf8 strings up to the maxlen.
+
+  Strings are matched character by character. For every non-wildcard
+  character there can be only one matching character in the matched string.
+
+  For a wild_one character ('_') any valid utf8 character will do. Below
+  numchars would mean a total number of vaid utf8 characters. It's a huge
+  number. A number of matching strings for wild_one will be numchars.
+
+  For a wild_many character ('%') any number of valid utf8 characters will do.
+  How many string will match it depends on the amount of non-wild_many
+  characters.  Say, if a number of non-wildcard characters is N, and a number
+  of wild_one characters is M, and the number of wild_many characters is K,
+  then for K=1 its wild_many character will match any number of valid utf8
+  characters from 0 to L=maxlen-N-M. The number of matching strings will be
+
+     1 + numchars + numchars^2 + numchars^3 + ... + numchars^L
+
+  Intermediate result: if M=K=0, the pattern will match only one string,
+  if M>0, K=0, the pattern will match numchars^M strings, if K=1, the
+  pattern will match
+
+     numchars^M + 1 + numchars + numchars^2 + ... + numchars^L
+
+  For a more visual notation, let's write these huge numbers not as
+  decimal or binary, but base numchars. Then the last number will be
+  a sum of two numbers: the first is one followed by M zeros, the second
+  constists of L+1 ones:
+
+    1000{...M...}000 + 111{...L+1...}1111
+
+  This could produce any of the following
+
+    111...112111...1111       if L > M, K = 1
+    100...001111...1111       if M > L, K = 1
+    2111111...111111111       if M = L, K = 1
+    1111111...111111111       if M = 0, K = 1
+    1000000...000000000       if K = 0, M > 0
+
+  There are two complications caused by multiple wild_many characters.
+  For, say, two wild_many characters, either can accept any number of utf8
+  characters, as long the the total amount of them is less then or equal to L.
+  Same logic applies to any number of non-consequent wild_many characters
+  (consequent wild_many characters count as one). This gives the number of
+  matching strings of
+
+    1 + F(K,1)*numchars + F(K,2)*numchars^2 + ... + F(K,L)*numchars^L
+
+  where F(K,R) is the "number of ways one can put R balls into K boxes",
+  that is C^{K-1}_{R+K-1}.
+
+  In the "base numchars" notation, it means that besides 0, 1, and 2,
+  an R-th digit can be F(K,R). For the purpose of comparison, we only need
+  to know the most significant digit, F(K, L).
+  While it can be huge, we don't need the exact value, it's a
+  a monotonously increasing function of K, so if K1>K2, F(K1,L) > F(K2,L)
+  and we can simply compare values of K instead of complex F(K,L).
+
+  The second complication: F(K,R) gives only an upper boundary, the
+  actual number of matched strings can be smaller.
+  Example: pattern "a%b%c" can match "abbc" as a(b)b()c, and as a()b(b)c.
+  F(2,1) = 2, but it's only one string "abbc".
+  We'll ignore it here under assumption that it almost never happens
+  in practice and this simplification won't noticeably disrupt the ordering.
+
+  The last detail: old get_sort function sorted by the non-wildcard prefix
+  length, so in "abc_" and "a_bc" the former one was sorted first. Strictly
+  speaking they're both equally specific, but to preserve the backward
+  compatible sorting we'll use the P "prefix length or 0 if no wildcards"
+  to break ties.
+
+  Now, let's compare two long numbers. Numbers are easy to compare,
+  the longer number is larger. If they both have the same lengths,
+  the one with the larger first digit is larger, and so on.
+
+  But there is no need to actually calculate these numbers.
+  Three numbers L, K, M (and P to break ties) are enough to describe a pattern
+  for a purpose of comparison. L/K/M triplets can be compared like this:
+
+  * case 1: if for both patterns L>M: compare L, K, M, in that order
+    because:
+      - if L1 > L2, the first number is longer
+      - If L1 == L2, then the first digit is a monotonously increasing function
+        of K, so the first digit is larger when K is larger
+      - if K1 == K2, then all other digits in these numbers would be the
+        same too, with the exception of one digit in the middle that
+        got +1 because of +1000{...M...}000. So, whatever number has a
+        larger M will get this +1 first.
+  * case 2: if for both patterns L<M: compare M, L, K, in that order
+  * case 3: if for both patterns L=M: compare L (or M), K
+  * case 4: if one L1>M1, other L2=M2: compare L, K, M
+  * case 5: if one L1<M1, other L2=M2: compare M, L, K
+  * case 6: if one pattern L1>M1, the other M2>L2: first is more generic
+     unless (case 6a) K1=K2=1,M1=0,M2=L2+1 (in that case - equal)
+
+  note that in case 3 one can use a rule from the case either 1 or 2,
+  in the case 4 one can use the rule from the case 1,
+  in the case 5 one can use the rule from the case 2.
+
+  for the case 6 and ignoring the special case 6a, to compare patterns by a
+  magic number as a function z(a,b,c), we must ensure that z(L1,K1,M1) is
+  greater than z(M2,L2,K2) when L1=M2. This can be done by an extra bit,
+  which is 1 for K and 0 for L. Thus, the magic number could be
+
+  case 1: (((L*2 + 1)*(maxlen+1) + K)*(maxlen+1) + M)*(maxlen+1) + P
+  case 2: ((M*2*(maxlen+1) + L)*(maxlen+1) + K)*(maxlen+1) + P
+
+  upper bound: L<=maxlen, M<=maxlen, K<=maxlen/2, P<maxlen
+  for a current maxlen=64, the magic number needs magic_bits bits.
+*/
+
+static ulonglong get_magic_sort(const char *templ, ...)
+{
+  ulonglong sort=0;
+  va_list args;
+  va_start(args, templ);
+
+  IF_DBUG(uint bits_used= 0,);
+
+  for (; *templ; templ++)
+  {
+    char *pat= va_arg(args, char*);
+
+    if (*templ == 'u')
+    {
+      /* Username. Can be empty (= anybody) or a literal. Encoded in one bit */
+      sort= (sort << 1) + !*pat;
+      IF_DBUG(bits_used++,);
+      continue;
+    }
+
+    /* A wildcard pattern.  Encoded in magic_bits bits.  */
+    uint maxlen= *templ == 'd' ? max_dbname_length : max_hostname_length;
+    DBUG_ASSERT(maxlen <= 255);
+    DBUG_ASSERT(*templ == 'd' || *templ == 'h');
+
+    uint N= 0, M= 0, K= 0, P= 0;
+    for (uint i=0; pat[i]; i++)
+    {
+      if (pat[i] == wild_many)
+      {
+        if (!K && !M) P= N;
+        K++;
+        while (pat[i+1] == wild_many) i++;
+        continue;
+      }
+      if (pat[i] == wild_one)
+      {
+        if (!K && !M) P= N;
+        M++;
+        continue;
+      }
+      if (pat[i] == wild_prefix && pat[i+1]) i++;
+      N++;
+    }
+
+    set_if_smaller(K, 31);
+    set_if_smaller(M, 31);
+
+    ulonglong L= K ? maxlen - N - M : 0, d= maxlen + 1, magic;
+    ulonglong d1= MY_MIN(d, 32);
+    if (L > M)
+      magic= (((L * 2 + 1) * d + K) * d1 + M) * d + P;
+    else
+      magic= (((M * 2 + 0) * d + L) * d1 + K) * d + P;
+    DBUG_ASSERT(magic < (1ULL << magic_bits));
+    sort= (sort << magic_bits) + magic;
+    IF_DBUG(bits_used+= magic_bits,);
+  }
+  DBUG_ASSERT(bits_used < 8*sizeof(sort));
+  va_end(args);
+  return ~sort;
+}
+#endif