Lambda expressions with List in C#

Question

Please consider this example List lst = new List { 1 2 3 4 }; int result = lst.Aggregate((x y) => x * y); As per the articles (xy) specifies input parameters and x*y as implementation my doubt is aggregate() is the method of list then it has to have inner implementation of aggregation by its own then what (x*y) specifies or rather needed here... Please mention the articles explaining lambda and list separately... Thanks in advance

Jignesh Trivedi · Answer

hi as per my understanding Enumerable.Aggregate is perform operation over a sequence. here if you perform multification it perform 1*2*3*4*5 = 120 Please refer http://msdn.microsoft.com/en-us/library/bb548651.aspx Aggregate is method of IEnumerable interface not List Class. hope this will help you.

Vulpes · Answer

With regard to: numbers.Sort((xy) => y-x); you're using the overload of the List .Sort method which takes a Comparison delegate as a parameter. This delegate is defined as follows: public delegate int Comparison (T x T y); So it takes two parameters of type T and returns an int. This is what the Lambda expression: (xy) => y-x does if x and y are ints i.e. numbers is a List . The Sort method uses this delegate to compare the elements of the array and since: if (x if (x > y) y - x is less than 0 if (x == y) y - x is zero it will be seen that it performs the sort in descending order. For example: var numbers = new List { 3 2 4 1 }; numbers.Sort((xy) => y-x); foreach(int i in numbers) Console.Write({0}  i); // 4 3 2 1

chaitanyapai pai · Answer

Aggregate is method of IEnumerable interface not List Class. thanks for this statement.

chaitanyapai pai · Answer

Let me make my question clear Ex: numbers.Sort((xy) => y-x); here what does y-x mean for sorting.. does (xy) => y-x return any value?