List<string> TrimStart elements

[sjoshi]

Is List.ConvertAll the only way to apply TrimStart to elements of a
List<stringor is there a better way ?
Currently I'm doing this...

List<stringlst = new List<string>();
lst.AddRange(new string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll<string>(delegate(string line) { return
line.TrimStart(null); });

thanks
Sunit

[Jon Skeet [C# MVP]]

sjoshi <sj****@ingr.comwrote:

Is List.ConvertAll the only way to apply TrimStart to elements of a
List<stringor is there a better way ?
Currently I'm doing this...

List<stringlst = new List<string>();
lst.AddRange(new string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll<string>(delegate(string line) { return
line.TrimStart(null); });

That's fine, yes. Nicer in C# 3, of course:

lst = lst.ConvertAll(line =line.TrimStart(null));

--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[Peter Duniho]

On 2007-11-02 13:18:22 -0700, sjoshi <sj****@ingr.comsaid:

Is List.ConvertAll the only way to apply TrimStart to elements of a
List<stringor is there a better way ?
Currently I'm doing this...

List<stringlst = new List<string>();
lst.AddRange(new string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll<string>(delegate(string line) { return
line.TrimStart(null); });

Since the String class is immutable, I think the List.ConvertAll()
method is a reasonably nice solution. You'll need _some_ sort of code
that replaces each list element with a new item, and that's pretty much
what ConvertAll() is for.

Other than an explicit for() loop indexing each list element, I don't
see any obvious alternative that is similarly compact and easy-to-read.

Is there something specific about List.ConvertAll() that you don't like?

Pete