I have this code:
List<int> list = new List<int>();
for (int i = 1; i <= n; i++)
list.Add(i);
Can I create this List<int> in one line with LINQ?
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3 Answers
List<int> list = Enumerable.Range(1, n).ToList();
Comments
List<int> list = Enumerable.Range(1, n).ToList();
Comments
If you need a lot of those lists, you might find the following extension method useful:
public static class Helper
{
public static List<int> To(this int start, int stop)
{
List<int> list = new List<int>();
for (int i = start; i <= stop; i++) {
list.Add(i);
}
return list;
}
}
Use it like this:
var list = 1.To(5);
Of course, for the general case, the Enumerable.Range thing the others posted may be more what you want, but I thought I'd share this ;) You can use this next time your Ruby-loving co-worker says how verbose C# is with that Enumerable.Range.
4 Comments
Pavel Minaev
If you're defining a general-purpose extension method like this, why not make it return
IEnumerable<int> rather than List<int>? The established convention for ranges pretty much everywhere is that they're lazy.
OregonGhost
Because it was asked for a List<int>, not for an IEnumerable<int>. I just wrote this for this question as a general sample how this could be shortened. And to show off to your Ruby co-worker, you would want to avoid the extra ToList call ;) Now of course, if you think this method is worth being a general-purpose method, go ahead and let it return IEnumerable<int>. This should even allow you to make it into an iterator block, instead of manually creating the list.
Adam Robinson
@Pavel: Because that's exactly what
Enumerable.Range does.Pavel Minaev
@Adam: I think the point here is mainly that
1.To(10) looks cuter than Enumerable.Range(1, 10). In Ruby, IIRC, to also returns a lazy sequence rather than eagerly filled collection.