Re: HashMap toString, what about the other way?

Supersedes: <map-keys-20071114231003@ram.dialup.fu-berlin.de>

Lasse Reichstein Nielsen <lrn@hotpop.com> writes:


  it also can handle keys that are maps itself.

  For example,

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.RoomSource;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { final RoomSource room = room( "< < a=b c=d >=1 >" );
    System.out.println( room.get( room( "< a=b c=d >" ))); }}

1

  Above, the map "< a=b c=d >" is used as a key and mapped to
  the value "1". "room.get" then retrieves the value "1" given
  the map "< a=b c=d >" as a key.

  The order or details of notation or redundant assertions do
  not matter in this case. Strings can also be written in
  brackets. Therefore, the notation "< c=d a = [b] c=d >" also
  is recognized as the same key, because it is the same map.

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.RoomSource;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { final RoomSource room = room( "< < a=b c=d >=1 >" );
    System.out.println( room.get( room( "< c=d a = [b] c=d >" ))); }}

1


  If a string value is to contain an equal sign or a space,
  it needs to be written in brackets. The routines will properly
  escape when serializing again. For example:

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.RoomSource;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { System.out.println( room( "< [abc=def ghi]=1 >" )); }}

< [abc=def ghi] =1 >

  The serialization routine has decided to use brackets here,
  because they are required. The space in front of the "=1" is a
  minor annoyance (formatting style bug), but it does not change
  the semantics just the style of the notation.

  If brackets are not needed, they will be omitted:

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.RoomSource;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { System.out.println( room( "< [abcdefghi]=1 >" )); }}

< abcdefghi =1 >

  Here, the serialization routine has detected that no
  bracket escape is needed.

  To see, how any string is escaped, you can insert it,
  and then serialize it. The following program adds the
  following two strings:

      abc[]def[ghi\jkl
      mno\]pqr\

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.Room;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { Room room = new Room();
    room.add( "abc[]def[ghi\\jkl" );
    room.add( "mno\\]pqr\\" );
    System.out.println( room ); }}

< [abc[]def\[ghi\jkl] [mno\\]pqr\|] >

  Somewhat unusual, but this is specified in this way,
  so that brackets and backslashes can be used inside a
  bracketed string without escaping as often as possible.
  Only unpaired brackets and backslashes at the end of a
  string needs special handling.

  This output now will be used as input to check, that it
  will be serialized again to the same output. Of course,
  Java now requires to double backslashes:

import java.lang.String;
import java.lang.System;
import de.dclj.ram.notation.unotal.Room;
import static de.dclj.ram.notation.unotal.RoomFromModule.room;

public final class Main
{ public static void main( final String argv[] )
  { Room room = room( "< [abc[]def\\[ghi\\jkl] [mno\\\\]pqr\\|] >" );
    System.out.println( room ); }}

< [abc[]def\[ghi\jkl] [mno\\]pqr\|] >

Supersedes: <map-keys-20071114231003@ram.dialup.fu-berlin.de>