Lambda expressions with List in C#

chaitanyapai pai

Please consider this example,

List lst = new List { 1, 2, 3, 4 };
int result = lst.Aggregate((x, y) => x * y);

As per the articles (x,y) 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,

4 Replies

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Vulpes13y ago
With regard to:

numbers.Sort((x,y) => 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: (x,y) => 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 < y) y - x is greater than 0

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((x,y) => y-x);
foreach(int i in numbers) Console.Write("{0} ", i); // 4 3 2 1
chaitanyapai pai13y ago
"Aggregate is method of IEnumerable interface not List Class." thanks for this statement.
chaitanyapai pai13y ago
Let me make my question clear, Ex: numbers.Sort((x,y) => y-x); here what does y-x mean for sorting.. does (x,y) => y-x return any value?
Jignesh Trivedi13y ago
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.