internal package Foswiki::Prefs::Stack

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UNPUBLISHED package Foswiki::Prefs::Stack

Foswiki preferences mechanism are like stacks:
  • Preferences pushed later have precedence over ones pushed earlier I must
  • be able to return (restore) to a state I was earlier

This stack can exist as an index, so preference data is not copied everywhere.

The index is composed by three elements:
  • A bitstring map. Each preference has a bitmap. Each bit corresponds to a level. The bit is 1 if the preference is defined at that level and 0 otherwise. If a preference is "defined" in some level, but it was finalized, then the corresponding bit is 0.
  • A level list storing a backend object that is associated with each level
  • A final hash that maps preferences to the level they were finalized.

This class deals with this stuff and must be used only by Foswiki::Prefs

ClassMethod new( $session )

Creates a new Stack object.

ObjectMethod finish()

Break circular references.

ObjectMethod size() → $size

Returns the size of the stack in number of levels.

ObjectMethod backAtLevel($level) → $back

Returns the backend object corresponding to $level. If $level is negative, consider that number from the top of the stack. -1 means the top element.

ObjectMethod finalizedBefore($pref, $level) → $boolean

Returns true if $pref was finalized somewhere earlier than $level. If $pref is finalized in $level or it's not finalized, returns true.

ObjectMethod finalized($pref) → $boolean

Returns true if $pref in finalized.

ObjectMethod prefs() → @prefs

Returns a list with the name of all defined prefs in the stack.

ObjectMethod prefIsDefined($pref) → $boolean

Returns true if $pref is defined somewhere in the stack.

ObjectMethod insert($type, $pref, $value) → $num

Define preference named $pref of type $type as $value. $type can be 'Local' or 'Set'.

Returns the number of inserted preferences (0 or 1).

ObjectMethod newLevel($back, $prefix)

Pushes all preferences in $back on the stack, except for the finalized ones. Optionally $prefix preferences name in the index. This feature is used by plugins: A preference PREF defined in MyPlugin topic should be referenced by MYPLUGIN_PREF. In this example $prefix is MYPLUGIN_.

ObjectMethod getDefinitionLevel($pref) → $level

Returns the $level in which $pref was defined or undef if it's not defined.

ObjectMethod getPreference($pref [, $level] ) → $value

Returns the $value of $pref, considering the stack rules (values in higher levels overrides those in lower levels).

Optionally consider preferences at most $level. This is usefull to get a preference of Web if the stack has Web/Subweb. This makes it possible to use the same stack for Web and Web/Subweb.

ObjectMethod clone($level ) → $stack

This constructs a new $stack object as a clone of this one, up to the given $level. If no $level is given, the resulting object is an extac copy.

ObjectMethod restore($level)

Restores tha stack to the state it was in the given $level.

Mathematical Considerations

by Gilmar Santos Jr, May 2009

The bitmap is built in an way to meet two properties:
  • It has the minimal possible length. (I)
  • If it exists in the hash, it has at least length 1. (II)

Preference levels 0-7 are in the first byte. 8-15 in the second and so on. If a preference is defined in levels 2 and 7, for example, its bitmap will have length 1, even if the stack is in 30th level. This is what minimal possible length means.

These two properties implies that the last byte of a bit string is non-zero.

Getting/Setting preferences with at most 8 levels

Let's consider the first scenario: at most 8 preference values. This means that bitmaps have one character. The built-in perl function ord converts a character to an integer between 0 and 255. If the character of a preference is 0, then the preference doesn't exist in the map hash, cause of the second listed property above.

This implies that I can always take the logarithm of ord($map). (III)

The question is: given a bitstring, what is the highest level containing a 1?

To answer this question let's consider the following mathematical expressions: (log2(X) means the logarithm of X in base 2)

log2(1) = 0; 1 == 1 * 2 ** 0; 1 in base 2 is "00000001" (considering one byte)
log2(2) = 1; 2 == 1 * 2 ** 1; 2 in base 2 is "00000010" (considering one byte)
log2(4) = 2; 4 == 1 * 2 ** 2; 4 in base 2 is "00000100" (considering one byte)
log2(8) = 3; 8 == 1 * 2 ** 3; 8 in base 2 is "00001000" (considering one byte)

Also notice that:

2 ** B <= X < 2 ** (B + 1) implies B <= log2(X) < (B + 1) 

This implies:

int(log2(X)) == B, for any X in the above rage.

Some examples:

int(log2(3)) = log2(2) = 1; 3 in base 2 is "00000011" (considering one byte)
int(log2(5)) = log2(4) = 2; 5 in base 2 is "00000101" (considering one byte)
int(log2(6)) = log2(4) = 2; 6 in base 2 is "00000110" (considering one byte)
int(log2(7)) = log2(4) = 2; 7 in base 2 is "00000111" (considering one byte)
int(log2(9)) = log2(8) = 3; 9 in base 2 is "00001001" (considering one byte)

The position of least significant bit is 0 and the position of the most significant bit is 8, then: int(log2(X)) is the position of the highest-significant bit equal to 1. This always holds. The complete mathematical proof is left as an exercise.

Back to the question what is the highest level containing a 1?

It's clear the answer is: int(log2(X)).

X is the number corresponding to the bitstring character, so X = ord($map).

Also,
log2(Y) == ln(Y)/ln(2), for any Y real positive

Then we have:
int(log2(X)) == int( ln( ord($map) ) / ln(2) )

Conclusion: considering (III) and at most 8 levels I can figure out in which level a preference is defined with the following O(1) operation:

$defLevel = int( ln( ord($map) ) / ln(2) );

Getting/Setting preferences with arbitrary number of levels

But preferences may have far more levels than 8. Now let's consider this general case. We'll reduce it to the at most 8 levels case.

At this point I must consider how perl built-in function vec works:
$a = '';
vec($a,  0, 1) = 1; print unpack("b*", $a);  # "10000000"
vec($a,  2, 1) = 1; print unpack("b*", $a);  # "10100000"
vec($a,  7, 1) = 1; print unpack("b*", $a);  # "10100001"
vec($a, 16, 1) = 1; print unpack("b*", $a);  # "1010000100000000100000000"

The least significant bit is the bit 0 of the first byte. The most significant bit is the bit 7 of the last byte. unpack with "b*" gives us this representation, that is different from the one we're used to, but it's only a representation. Test for yourself:

$a = '';
vec($a, 0, 1) = 1; print ord($a); #   1
vec($a, 2, 1) = 1; print ord($a); #   5
vec($a, 7, 1) = 1; print ord($a); # 133

Since ord() operates with one character (or with the first one, if length($a) > 1), we have to figure out a way to deal with preferences bigger than 8 levels.

The level to consider in order to get a preference value is the highest in which it was defined. Because of properties (I) and (II) above, this level is in the last byte of the bitmap. This implies that no matter the value of the other bytes are, I need to consider only the last byte. (IV)

Since (IV) holds, we can reduce the general case to the restricted case as follows: we calculate the level considering the last byte. We'll get $L in [0,7]. Then we transform this value to the correct, considering that the bit 0 of the last byte is the bit (N - 1) * 8 of the general string, where N is the total number of bytes. Examples:

1  byte: bit 0 of the last byte is bit (1 - 1) * 8 == 0  of the string
2 bytes: bit 0 of the last byte is bit (2 - 1) * 8 == 8  of the string
3 bytes: bit 0 of the last byte is bit (3 - 1) * 8 == 16 of the string

and so on.

So, considering the general case where $map has arbitrary length, the general answer to what is the highest level containing a 1? is:

$defLevel = int( log( ord( substr($map, -1) ) ) / ln(2) ) + 
            (length($map) - 1) * 8;

substr($map, -1) is the last byte of =$map= and because of (I) it's non-zero, so log( ord( substr($map,-1) ) ) exists. Because of (II), length($map) is at least 1. So this general expression is always valid.

Growth/Shrink operations with at most 8 levels

There are growth and shrink operations on the stack and hence in the bitmaps. These operations must keep (I) and (II). Let's consider the initial case: $stack->{map} is an empty hashref, so both (I) and (II) holds.

The addition of a preference uses vec(), that expands the string as (and only as) needed, so (I) holds. And if the preference is being added, then it must exist in preferences map, so (II) also holds.

The restore operation is more complex: if we're restoring to level L, this means that all bits above level L must be 0. I can accomplish this using bitwise AND (&):

Considering at most 8 bytes, let's assume we want to restore to the level 5. Notice that:

2 ** (5 + 1) == 64 == "01000000"
64 - 1 == 63 == "00111111"

Bits 0-5 are 1 and all others are 0.

And Since:

(1 & X) == X
(0 & X) == 0

we can build a mask using this process and apply it to the map and we'll get the bitmap restored to the desired level. So, in order to restore to level $L we build a mask as ((2 ** (L+1)) - 1) and perform:

$map &= $mask;

If the result is 0, we need to remove that preference from the hash, so both (I) and (II) holds.

Growth/Shrink operations with arbitrary number of levels

Now considering the general case: if we want to restore to level $L, we need to build a mask whose bits 0-L are 1. This mask will have int($L/8) + 1 bytes.

0  <= $L <  8 implies the mask 1-byte  long.
8  <= $L < 16 implies the mask 2-bytes long.
16 <= $L < 24 implies the mask 3-bytes long. 

and so on. We conclude that all bytes of the mask, except the last, will be \xFF (all bits 1). If we map $L to [0,7], then we have the restricted case above.

The number of bytes except the last in the bitstring is int($L/8). The bit position of $L in the last byte is ($L % 8):

Level  8 corresponds to bit 0 of the second byte. int(8/8)  = 1.  8 % 8 = 0.
Level  9 corresponds to bit 1 of the second byte. int(9/8)  = 1.  9 % 8 = 1.
Level 15 corresponds to bit 7 of the second byte. int(15/8) = 1. 15 % 8 = 7.
Level 16 corresponds to bit 0 of the third byte.  int(16/8) = 2. 16 % 8 = 0.

So the general way to build the mask is:

$mask = ("\xFF" x int($L/8)); # All bytes except the last have all bits 1.
$mask .= chr( ( 2**( ( $L % 8 ) + 1 ) ) - 1 ); # The last byte is built based
                                               # on the restricted case above.

The $mask has the minimal possible length, cause the way it's built. $map & $mask has at most length($mask) bytes, cause the way & works. But we still must guarantee (I) and (II), so we need to purge the possible zero-bytes in the end of the bitstring:

while (ord(substr($map, -1)) == 0 && length($map) > 0 ) {
    substr($map, -1) = '';
}

We need to test if length($map) is greater than 0, otherwise we may enter on an infinite loop, if all bytes of the result are 0.

This while guarantee (I) above. Then we check if the resulting $map has length 0. If so we remove the pref from the hash, so (II) is also achieved.

Other Considerations

This implementation is more complex than the "natural" way, but using this:
  • We avoid to have more than one copy of preference values
  • This architecture (index separated from the values) make it easy to change the way values are stored.

Also, consider it's slow to copy large chunks of data around. All copied values in this architecture are far smaller than the preferences values (a typical big bitstring has less than 4 bytes, while a preference value is bigger than this).

pack and unpack are not used cause they are not needed and cause the way to know the level where a preference is defined is an O(1) operation that depends on the packed string.

Topic revision: r1 - 03 Feb 2016, UnknownUser
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