How can I access the next value in a collection inside a foreach loop in C#?

Question

I'm working in C# and with a sorted List<T> of structs. I'm trying to iterate through the List and for each iteration I'd like to access the next member of the list. Is there a way to do this?

Pseudocode example:

foreach (Member member in List)
{
    Compare(member, member.next);
}

Answer 1

You can't. Use a for instead

for(int i=0; i<list.Count-1; i++)
   Compare(list[i], list[i+1]);

Answer 2

Use a regular for loop with an index, and compare list[i] and list[i+1]. (But make sure to only loop until the second-to-last index.)

Or, if you really want to use a foreach, you can keep a Member reference to the previous member and check the next time around. But I wouldn't recommend it.

Answer 3

You could just keep the previous value instead:

T prev = default(T);
bool first = true;
foreach(T item in list) {
    if(first) {
        first = false;
    } else {
        Compare(prev, item);
    }
    prev = item;
}

Answer 4

LINQ might be your friend here. This approach will work with anything that's IEnumerable<T>, not just IList<T> collections, which is very useful if your collection never ends or is otherwise calculated on-the-fly:

class Program {
    static void Main(string[] args) {
        var list = new List<Int32> { 1, 2, 3, 4, 5 };
        foreach (var comparison in list.Zip(list.Skip(1), Compare)) {
            Console.WriteLine(comparison);
        }
        Console.ReadKey();
    }

    static Int32 Compare(Int32 first, Int32 second) {
        return first - second;
    }
}

Answer 5

If one were so inclined, you could probably write an Extension method for this as well...

public static void ForEachNext<T>(this IList<T> collection, Action<T, T> func)
{
    for (int i = 0; i < collection.Count - 1; i++)
        func(collection[i], collection[i + 1]);
}

Usage:

List<int> numList = new List<int> { 1, 3, 5, 7, 9, 11, 13, 15 };

numList.ForEachNext((first, second) => 
{
    Console.WriteLine(string.Format("{0}, {1}", first, second));
});