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(ne w string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll< string>(delegat e(string line) { return
line.TrimStart( null); });

thanks
Sunit

[Jon Skeet [C# MVP]]

Re: List&lt;string& gt; TrimStart elements

sjoshi <sjoshi@ingr.co mwrote:

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(ne w string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll< string>(delegat e(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 - <skeet@pobox.co m>
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]

Re: List&lt;string& gt; TrimStart elements

On 2007-11-02 13:18:22 -0700, sjoshi <sjoshi@ingr.co msaid:

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(ne w string[] {"A test", " that you", " need to see!" });
lst = lst.ConvertAll< string>(delegat e(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