objective c - How should I construct an NSDictionary with multiple keys? -

I have a situation where I want to map a pair of items for a dictionary of information. I would like to avoid creating an NSDictionary of NSDictionaries of NSDictionaries because it is only confusing and difficult to maintain. For example, if I have two squares, foo and bar, let me know {Foo, Bar} -> {NSDictionary}

Want to have a choice other than just creating a custom class (Can only be used as a dictionary key on the basis of two types of question? Probably like the pair of STL Something that I'm not right now As you said, you can create a pair of classes:

 
 
  // Pair.h @ Interface pair: NSPacked & lt; encoding & gt; {ID & lt; encoding> left; ID & lt; NSCopying & gt; Correct;} @ Property (Nonomatic, Readonly) ID & lt; NScoping & gt; Left; @ Property (Nonomatic, Readonly) ID & lt; NScoping; Right; + (ID) pair with lift: (ID & lt; NSCPIP & gt;) L Correct: & Lt; nspip & gt; R; - (ID ) InitWithLeft: (ID & lt; Ansseepiaipi & gt;) L Right (ID & lt; NCPPing & gt;) R; @ And // pir.m # Imports "Pair H." @Epilmentation pair @ Synthesize left, right; + (ID) pair with lift: (id & lieutenant; nscpijg) l correct: (id & lieutenant; ncpip> gt; r [return [[sg class]] allok with input: lift: r ] AutoWrecks]; } - (id) initWithLeft: (ID & Lt; NSCPIG & gt;) El Correct: (id & lt; NCPPI) R {if (self = [super init]) {left = [l copy]; Right = [r copy]; } Healthy return; } - (zero) finally {[left release], left = zero; [Right release], right = zero; [Super final form]; } - (zero) Delok [[left release], left = zero; [Right release], right = zero; [Super DeLoc]; } - (ID) Copywritz zone: (NSZEN *) region {pair} * copy = [[self class] incoming] initWithLeft: [self remaining] true: [self correct]]; Return copy; } - (BOOL) is aqual: (id) other {if ([Other Hackandoff class: [pair class]] == no {return not; } Return ([[Self left] is Acqual: [other left]] & [[Self correct] is Acqual: [Other rights]]); } - (NSUntengar) hash (probably probably "hashish") is not enough, but possibly sufficient enough return [[self left] hash] + [[self right] hash]; } @end  

Edit:

Some notes on the making of such things which can be keys:

1. call them Encode the protocol , since we want us to change the key from the down side.
2. Since the pair has been copied, I'm sure the objects are also copied in the pair.
3. Keys have to be executed is Excel: . I apply the hash method to good measure, but it is probably not necessary.