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lib/Algorithm/Diff.pm  view on Meta::CPAN

        diff sdiff compact_diff
        traverse_sequences traverse_balanced );

    @lcs    = LCS( \@seq1, \@seq2 );
    $lcsref = LCS( \@seq1, \@seq2 );
    $count  = LCS_length( \@seq1, \@seq2 );

    ( $seq1idxref, $seq2idxref ) = LCSidx( \@seq1, \@seq2 );


    # Complicated interfaces:

    @diffs  = diff( \@seq1, \@seq2 );

    @sdiffs = sdiff( \@seq1, \@seq2 );

    @cdiffs = compact_diff( \@seq1, \@seq2 );

    traverse_sequences(
        \@seq1,
        \@seq2,
        {   MATCH     => \&callback1,
            DISCARD_A => \&callback2,
            DISCARD_B => \&callback3,
        },
        \&key_generator,
        @extra_args,
    );

    traverse_balanced(
        \@seq1,
        \@seq2,
        {   MATCH     => \&callback1,
            DISCARD_A => \&callback2,
            DISCARD_B => \&callback3,
            CHANGE    => \&callback4,
        },
        \&key_generator,
        @extra_args,
    );


=head1 INTRODUCTION

(by Mark-Jason Dominus)

I once read an article written by the authors of C<diff>; they said
that they worked very hard on the algorithm until they found the
right one.

I think what they ended up using (and I hope someone will correct me,
because I am not very confident about this) was the `longest common
subsequence' method.  In the LCS problem, you have two sequences of
items:

    a b c d f g h j q z

    a b c d e f g i j k r x y z

and you want to find the longest sequence of items that is present in
both original sequences in the same order.  That is, you want to find
a new sequence I<S> which can be obtained from the first sequence by
deleting some items, and from the second sequence by deleting other
items.  You also want I<S> to be as long as possible.  In this case I<S>
is

    a b c d f g j z

From there it's only a small step to get diff-like output:

    e   h i   k   q r x y
    +   - +   +   - + + +

This module solves the LCS problem.  It also includes a canned function
to generate C<diff>-like output.

It might seem from the example above that the LCS of two sequences is
always pretty obvious, but that's not always the case, especially when
the two sequences have many repeated elements.  For example, consider

    a x b y c z p d q
    a b c a x b y c z

A naive approach might start by matching up the C<a> and C<b> that
appear at the beginning of each sequence, like this:

    a x b y c         z p d q
    a   b   c a b y c z

This finds the common subsequence C<a b c z>.  But actually, the LCS
is C<a x b y c z>:

          a x b y c z p d q
    a b c a x b y c z

or

    a       x b y c z p d q
    a b c a x b y c z

=head1 USAGE

(See also the README file and several example
scripts include with this module.)

This module now provides an object-oriented interface that uses less
memory and is easier to use than most of the previous procedural
interfaces.  It also still provides several exportable functions.  We'll
deal with these in ascending order of difficulty:  C<LCS>,
C<LCS_length>, C<LCSidx>, OO interface, C<prepare>, C<diff>, C<sdiff>,
C<traverse_sequences>, and C<traverse_balanced>.

=head2 C<LCS>

Given references to two lists of items, LCS returns an array containing
their longest common subsequence.  In scalar context, it returns a
reference to such a list.

    @lcs    = LCS( \@seq1, \@seq2 );
    $lcsref = LCS( \@seq1, \@seq2 );

lib/Algorithm/Diff.pm  view on Meta::CPAN

    $obj        = Algorithm::Diff->new( \@seq1, \@seq2, [ \%opts ] );
    $pos        = $obj->Next(  [ $count : 1 ] );
    $revPos     = $obj->Prev(  [ $count : 1 ] );
    $obj        = $obj->Reset( [ $pos : 0 ] );
    $copy       = $obj->Copy(  [ $pos, [ $newBase ] ] );
    $oldBase    = $obj->Base(  [ $newBase ] );

Note that all of the following methods C<die> if used on an object that
is "reset" (not currently pointing at any hunk).

    $bits       = $obj->Diff(  );
    @items|$cnt = $obj->Same(  );
    @items|$cnt = $obj->Items( $seqNum );
    @idxs |$cnt = $obj->Range( $seqNum, [ $base ] );
    $minIdx     = $obj->Min(   $seqNum, [ $base ] );
    $maxIdx     = $obj->Max(   $seqNum, [ $base ] );
    @values     = $obj->Get(   @names );

Passing in C<undef> for an optional argument is always treated the same
as if no argument were passed in.

=item C<Next>

    $pos = $diff->Next();    # Move forward 1 hunk
    $pos = $diff->Next( 2 ); # Move forward 2 hunks
    $pos = $diff->Next(-5);  # Move backward 5 hunks

C<Next> moves the object to point at the next hunk.  The object starts
out "reset", which means it isn't pointing at any hunk.  If the object
is reset, then C<Next()> moves to the first hunk.

C<Next> returns a true value iff the move didn't go past the last hunk.
So C<Next(0)> will return true iff the object is not reset.

Actually, C<Next> returns the object's new position, which is a number
between 1 and the number of hunks (inclusive), or returns a false value.

=item C<Prev>

C<Prev($N)> is almost identical to C<Next(-$N)>; it moves to the $Nth
previous hunk.  On a 'reset' object, C<Prev()> [and C<Next(-1)>] move
to the last hunk.

The position returned by C<Prev> is relative to the I<end> of the
hunks; -1 for the last hunk, -2 for the second-to-last, etc.

=item C<Reset>

    $diff->Reset();     # Reset the object's position
    $diff->Reset($pos); # Move to the specified hunk
    $diff->Reset(1);    # Move to the first hunk
    $diff->Reset(-1);   # Move to the last hunk

C<Reset> returns the object, so, for example, you could use
C<< $diff->Reset()->Next(-1) >> to get the number of hunks.

=item C<Copy>

    $copy = $diff->Copy( $newPos, $newBase );

C<Copy> returns a copy of the object.  The copy and the original object
share most of their data, so making copies takes very little memory.
The copy maintains its own position (separate from the original), which
is the main purpose of copies.  It also maintains its own base.

By default, the copy's position starts out the same as the original
object's position.  But C<Copy> takes an optional first argument to set the
new position, so the following three snippets are equivalent:

    $copy = $diff->Copy($pos);

    $copy = $diff->Copy();
    $copy->Reset($pos);

    $copy = $diff->Copy()->Reset($pos);

C<Copy> takes an optional second argument to set the base for
the copy.  If you wish to change the base of the copy but leave
the position the same as in the original, here are two
equivalent ways:

    $copy = $diff->Copy();
    $copy->Base( 0 );

    $copy = $diff->Copy(undef,0);

Here are two equivalent way to get a "reset" copy:

    $copy = $diff->Copy(0);

    $copy = $diff->Copy()->Reset();

=item C<Diff>

    $bits = $obj->Diff();

C<Diff> returns a true value iff the current hunk contains items that are
different between the two sequences.  It actually returns one of the
follow 4 values:

=over 4

=item 3

C<3==(1|2)>.  This hunk contains items from @seq1 and the items
from @seq2 that should replace them.  Both sequence 1 and 2
contain changed items so both the 1 and 2 bits are set.

=item 2

This hunk only contains items from @seq2 that should be inserted (not
items from @seq1).  Only sequence 2 contains changed items so only the 2
bit is set.

=item 1

This hunk only contains items from @seq1 that should be deleted (not
items from @seq2).  Only sequence 1 contains changed items so only the 1
bit is set.

=item 0

This means that the items in this hunk are the same in both sequences.
Neither sequence 1 nor 2 contain changed items so neither the 1 nor the
2 bits are set.

=back

=item C<Same>

C<Same> returns a true value iff the current hunk contains items that
are the same in both sequences.  It actually returns the list of items
if they are the same or an empty list if they aren't.  In a scalar
context, it returns the size of the list.

=item C<Items>

    $count = $diff->Items(2);
    @items = $diff->Items($seqNum);