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// Now calculate the factorial and display it. // av == aggregated value, e == element int agg = intSequence.Aggregate((av, e) => av * e); Console.WriteLine("{0}! = {1}", N, agg); In the previous code, I generate a sequence containing the integers from 1 to 5 using the Range operator. After displaying each element in the generated sequence, I call the Aggregate operator passing a lambda expression that multiplies the passed aggregated value with the passed element itself. The following are the results: 1 2 3 4 5 5! = 120

ILookup<string, string> lookup = Actor2.GetActors() .ToLookup(k => k.birthYear, a => string.Format("{0} {1}", a.firstName, a.lastName), new MyStringifiedNumberComparer()); // Let's see if we can find the 'one' born in 1964. IEnumerable<string> actors = lookup["0001964"]; foreach (var actor in actors) Console.WriteLine("{0}", actor); Here is the output: Keanu Reeves Sandra Bullock You can see that we index into the Lookup using a key value different from either of the values retrieved using that key, so we can tell our equality comparison object is working. And instead of storing the entire Actor2 object, we merely store the string we are interested in.

... <div title="Date Field Component" > <input name="dateField" value="26 January 2005" > <img class="overlay" src="inputDateOverlay.gif" > </div>

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Caution You should be careful when using this version of the Aggregate operator that the first element doesn t get operated on twice, since it is passed in as the input value and the element for the first element. In the previous example, my first call to my func lambda expression would have passed in 1 and 1. Since I just multiplied these two values, and they are both ones, there is no bad side effect. But if I had added the two values, I would have a sum that included the first element twice.

public static bool SequenceEqual<T>( this IEnumerable<T> first, IEnumerable<T> second); This operator enumerates each input sequence in parallel, comparing the elements of each using the System.Object.Equals method. If the elements are all equal and the sequences have the same number of elements, the operator returns true. Otherwise, it returns false. The second prototype of the operator works just as the first, except an IEqualityComparer<T> comparer object can be used to determine element equality.

public static bool SequenceEqual<T>( this IEnumerable<T> first, IEnumerable<T> second, IEqualityComparer<T> comparer);

For the second prototype s example, shown in Listing 5-64, I roll my own version of the Sum operator. Listing 5-64. An Example of the Second Aggregate Prototype IEnumerable<int> intSequence = Enumerable.Range(1, 10); // I'll just output the sequence so all can see it. foreach (int item in intSequence) Console.WriteLine(item); Console.WriteLine("--"); // Now calculate the sum and display it. int sum = intSequence.Aggregate(0, (s, i) => s + i); Console.WriteLine(sum); Notice that I passed 0 as the seed for this call to the Aggregate operator. And the envelope please: 1 2 3 4 5 6 7 8 9 10 -55

string[] presidents = { "Adams", "Arthur", "Buchanan", "Bush", "Carter", "Cleveland", "Clinton", "Coolidge", "Eisenhower", "Fillmore", "Ford", "Garfield", "Grant", "Harding", "Harrison", "Hayes", "Hoover", "Jackson", "Jefferson", "Johnson", "Kennedy", "Lincoln", "Madison", "McKinley", "Monroe", "Nixon", "Obama", "Pierce", "Polk", "Reagan", "Roosevelt", "Taft", "Taylor", "Truman", "Tyler", "Van Buren", "Washington", "Wilson"}; bool eq = presidents.SequenceEqual(presidents);

Code Sample 2-2. Parameterized HTML for the Input Date Component <style type="text/css" > .overlay { position:relative; left:-10px; bottom:-10px; } </style> ... <div title="[title]"> <input name="[clientId]" value="[converted value]" > <img class="overlay" src="inputDateOverlay.gif" > </div> In Code Sample 2-2, you map the HTML attributes to their corresponding UIComponent attributes that are used during rendering.

As you can see, I got the exact same results that I did when calling the Sum operator in Listing 5-51.

Console.WriteLine(eq); And here are the results: True That seems a little cheap, doesn t it OK, we will make it a little more difficult, as shown in Listing 512.

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