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copied!<p>I am not sure what you're really asking here, but let's see if any of the following helps.</p> <p>First consider this method based on your code fragment:</p> <pre><code>- (void) test { NSMutableArray *Array1 = [NSMutableArray new]; NSMutableArray *Array2 = [NSMutableArray new]; [Array1 addObject:@"Hello"]; [Array1 addObject:@"There"]; [Array2 addObject:@"Hi"]; [Array2 addObject:@"Welcome"]; NSMutableDictionary *dictionaryOfArrays = [NSMutableDictionary new]; [dictionaryOfArrays setObject:Array2 forKey:Array1]; // arrays are ordered, use key with elements in different order to show this [dictionaryOfArrays setObject:Array2 forKey:@[@"There", @"Hello"]]; NSLog(@"dictionary: %@", dictionaryOfArrays); } </code></pre> <p>This will output to the console:</p> <pre><code>dictionary: { ( There, Hello ) = ( Hi, Welcome ); ( Hello, There ) = ( Hi, Welcome ); } </code></pre> <p>The shows that the <em>key</em> can be an array - and the order of elements in the array is important.</p> <p>However do you really want to use arrays as keys? While having the <em>values</em> as arrays is common, it is not common for the key to be an array as well.</p> <p>You say you are getting either <code>( )</code> or "null". Well the first of those could be an array with no elements - which is distinct from no array. As to the second, you cannot store <code>nil</code> in a <code>NSDictionary</code> but if you look up a value for a non-existant key then <code>nil</code> is returned - i.e. a <code>nil</code> return means a non-existant key.</p> <p>Finally in case it is relevant, when a key/value pair is added to a dictionary the key is <em>copied</em> while the value is simply <em>referenced</em>. This means that if you use a mutable object as a key, e.g. a <code>NSMutableString</code> instance, then subsequent mutations of the object do not effect the dictionary. However if the value (of the key/value pair) is a mutable object then a lookup operation returns the exact same object as was inserted, which could have been mutated since insertion.</p> <p>The code you've added to your question doesn't clarify at all.</p> <p>Hope something here helps.</p>

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