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Lambda Expression using Foreach Clause [duplicate]

Possible Duplicate:
Why is there not a ForEach extension method on the IEnumerable interface?

EDIT

For reference, here's the blog post which eric referred to in the comments

https://ericlippert.com/2009/05/18/foreach-vs-foreach/

ORIG

More of a curiosity I suppose but one for the C# Specification Savants...

Why is it that the ForEach() clause doesn't work (or isn't available) for use on IQueryable/IEnumerable result sets...

You have to first convert your results ToList() or ToArray() Presumably theres a technical limitation to the way C# iterates IEnumerables Vs. Lists... Is it something to do with the Deferred Execution's of IEnumerables/IQuerable Collections. e.g.

var userAgentStrings = uasdc.UserAgentStrings
    .Where<UserAgentString>(p => p.DeviceID == 0 && 
                            !p.UserAgentString1.Contains("msie"));
//WORKS            
userAgentStrings.ToList().ForEach(uas => ProcessUserAgentString(uas));         

//WORKS
Array.ForEach(userAgentStrings.ToArray(), uas => ProcessUserAgentString(uas));

//Doesn't WORK
userAgentStrings.ForEach(uas => ProcessUserAgentString(uas));
like image 707
Eoin Campbell Avatar asked May 13 '09 16:05

Eoin Campbell


1 Answers

It's perfectly possible to write a ForEach extension method for IEnumerable<T>.

I'm not really sure why it isn't included as a built-in extension method:

  • Maybe because ForEach already existed on List<T> and Array prior to LINQ.
  • Maybe because it's easy enough to use a foreach loop to iterate the sequence.
  • Maybe because it wasn't felt to be functional/LINQy enough.
  • Maybe because it isn't chainable. (It's easy enough to make a chainable version that yields each item after performing an action, but that behaviour isn't particularly intuitive.)

public static void ForEach<T>(this IEnumerable<T> source, Action<T> action)
{
    if (source == null) throw new ArgumentNullException("source");
    if (action == null) throw new ArgumentNullException("action");

    foreach (T item in source)
    {
        action(item);
    }
}
like image 112
LukeH Avatar answered Nov 10 '22 08:11

LukeH