Let's say I have a class
public class MyObject
{
public int SimpleInt{get;set;}
}
And I have a List<MyObject>, and I ToList() it and then change one of the SimpleInt, will my change be propagated back to the original list. In other words, what would be the output of the following method?
public void RunChangeList()
{
var objs = new List<MyObject>(){new MyObject(){SimpleInt=0}};
var whatInt = ChangeToList(objs );
}
public int ChangeToList(List<MyObject> objects)
{
var objectList = objects.ToList();
objectList[0].SimpleInt=5;
return objects[0].SimpleInt;
}
Why?
P/S: I'm sorry if it seems obvious to find out. But I don't have compiler with me now...
I don't see anywhere in the documentation that ToList() is always guaranteed to return a new list. If an IEnumerable is a List, it may be more efficient to check for this and simply return the same List.
The worry is that sometimes you may want to be absolutely sure that the returned List is != to the original List. Because Microsoft doesn't document that ToList will return a new List, we can't be sure (unless someone found that documentation). It could also change in the future, even if it works now.
new List(IEnumerable enumerablestuff) is guaranteed to return a new List. I would use this instead.
Just stumble upon this old post and thought of adding my two cents. Generally, if I am in doubt, I quickly use the GetHashCode() method on any object to check the identities. So for above -
public class MyObject
{
public int SimpleInt { get; set; }
}
class Program
{
public static void RunChangeList()
{
var objs = new List<MyObject>() { new MyObject() { SimpleInt = 0 } };
Console.WriteLine("objs: {0}", objs.GetHashCode());
Console.WriteLine("objs[0]: {0}", objs[0].GetHashCode());
var whatInt = ChangeToList(objs);
Console.WriteLine("whatInt: {0}", whatInt.GetHashCode());
}
public static int ChangeToList(List<MyObject> objects)
{
Console.WriteLine("objects: {0}", objects.GetHashCode());
Console.WriteLine("objects[0]: {0}", objects[0].GetHashCode());
var objectList = objects.ToList();
Console.WriteLine("objectList: {0}", objectList.GetHashCode());
Console.WriteLine("objectList[0]: {0}", objectList[0].GetHashCode());
objectList[0].SimpleInt = 5;
return objects[0].SimpleInt;
}
private static void Main(string[] args)
{
RunChangeList();
Console.ReadLine();
}
And answer on my machine -
So essentially the object that list carries remain the same in above code. Hope the approach helps.
var objectList = objects.ToList();
objectList[0].SimpleInt=5;
This will update the original object as well. The new list will contain references to the objects contained within it, just like the original list. You can change the elements either and the update will be reflected in the other.
Now if you update a list (adding or deleting an item) that will not be reflected in the other list.
In the case where the source object is a true IEnumerable (i.e. not just a collection packaged an as enumerable), ToList() may NOT return the same object references as in the original IEnumerable. It will return a new List of objects, but those objects may not be the same or even Equal to the objects yielded by the IEnumerable when it is enumerated again
A new list is created but the items in it are references to the orginal items (just like in the original list). Changes to the list itself are independent, but to the items will find the change in both lists.
ToList will always create a new list, which will not reflect any subsequent changes to the collection.
However, it will reflect changes to the objects themselves (Unless they're mutable structs).
In other words, if you replace an object in the original list with a different object, the ToList will still contain the first object.
However, if you modify one of the objects in the original list, the ToList will still contain the same (modified) object.
Yes, it creates a new list. This is by design.
The list will contain the same results as the original enumerable sequence, but materialized into a persistent (in-memory) collection. This allows you to consume the results multiple times without incurring the cost of recomputing the sequence.
The beauty of LINQ sequences is that they are composable. Often, the IEnumerable<T> you get is the result of combining multiple filtering, ordering, and/or projection operations. Extension methods like ToList() and ToArray() allow you to convert the computed sequence into a standard collection.
The accepted answer correctly addresses the OP's question based on his example. However, it only applies when ToList is applied to a concrete collection; it does not hold when the elements of the source sequence have yet to be instantiated (due to deferred execution). In case of the latter, you might get a new set of items each time you call ToList (or enumerate the sequence).
Here is an adaptation of the OP's code to demonstrate this behaviour:
public static void RunChangeList()
{
var objs = Enumerable.Range(0, 10).Select(_ => new MyObject() { SimpleInt = 0 });
var whatInt = ChangeToList(objs); // whatInt gets 0
}
public static int ChangeToList(IEnumerable<MyObject> objects)
{
var objectList = objects.ToList();
objectList.First().SimpleInt = 5;
return objects.First().SimpleInt;
}
Whilst the above code may appear contrived, this behaviour can appear as a subtle bug in other scenarios. See my other example for a situation where it causes tasks to get spawned repeatedly.
ToList will create a brand new list.
If the items in the list are value types, they will be directly updated, if they are reference types, any changes will be reflected back in the referenced objects.
I think that this is equivalent to asking if ToList does a deep or shallow copy. As ToList has no way to clone MyObject, it must do a shallow copy, so the created list contains the same references as the original one, so the code returns 5.
From the Reflector'd source:
public static List<TSource> ToList<TSource>(this IEnumerable<TSource> source)
{
if (source == null)
{
throw Error.ArgumentNull("source");
}
return new List<TSource>(source);
}
So yes, your original list won't be updated (i.e. additions or removals) however the referenced objects will.
Source: Stackoverflow.com