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High memory consumption with Enumerable.Range?

Originally i wanted to know whether ToList allocates more memory than using the constructor of List<T> which takes an IEnumerable<T> (no difference).

For test purposes i used Enumerable.Range to create a source array that i could use to create an instance of List<int> via 1.ToList and 2.constructor. Both are creating copies.

This is how I came to notice a great difference in memory consumption between:

  1. Enumerable.Range(1, 10000000) or
  2. Enumerable.Range(1, 10000000).ToArray()

When i use the first and call ToList the resulting object ineeds ~60% more memory than the Array(38,26MB/64MB).

Q: What is the reason for this or where is my error in reasoning?

var memoryBefore = GC.GetTotalMemory(true);
var range = Enumerable.Range(1, 10000000);
var rangeMem = GC.GetTotalMemory(true) - memoryBefore; // negligible
var list = range.ToList();
var memoryList = GC.GetTotalMemory(true) - memoryBefore - rangeMem;

String memInfoEnumerable = String.Format("Memory before: {0:N2} MB List: {1:N2} MB"
    , (memoryBefore / 1024f) / 1024f
    , (memoryList   / 1024f) / 1024f);
// "Memory before: 0,11 MB List: 64,00 MB"

memoryBefore = GC.GetTotalMemory(true);
var array = Enumerable.Range(1, 10000000).ToArray();
var memoryArray = GC.GetTotalMemory(true) - memoryBefore;
list = array.ToList();
memoryList = GC.GetTotalMemory(true) - memoryArray;

String memInfoArray = String.Format("Memory before: {0:N2} MB Array: {1:N2} MB List: {2:N2} MB"
   , (memoryBefore / 1024f) / 1024f
   , (memoryArray  / 1024f) / 1024f
   , (memoryList   / 1024f) / 1024f);
// "Memory before: 64,11 MB Array: 38,15 MB List: 38,26 MB"
like image 814
Tim Schmelter Avatar asked May 09 '12 15:05

Tim Schmelter


2 Answers

This probably relates to the doubling algorithm used to resize the backing buffer when adding to a list. When you allocate as an array, the length of that is known, and can be queried by checking for IList[<T>] and/or ICollection[<T>]; thus it can allocate a single array, right-sized the first time, and then just block-copy the contents.

With the sequence this is not possible (the sequence does not expose the length in any accessible way); thus it must instead fall back to "keep filling up the buffer; if full, double it and copy".

Obviously this needs approx double the memory.

An interesting test would be:

var list = new List<int>(10000000);
list.AddRange(Enumerable.Range(1, 10000000));

This will allocate the right size initially, while still using the sequence.

tl;dr; the constructor, when passed a sequence, first checks to see if it can obtain the length by casting to a well-known interface.

like image 163
Marc Gravell Avatar answered Oct 17 '22 03:10

Marc Gravell


It's because of the doubling algorithm used to create the backing array in a List. IEnumerable has no Count property so it can't pre-allocate the backing array to be the target size when you call ToList. In fact, on each call to MoveNext you are calling a corresponding Add on the List.

However Array.ToList can override the base ToList behaviour to initialise the list to the correct capacity. Also, it could be the List in it's constructor which tries to downcast the IEnumerable reference it has to known collection types such as IList, ICollection, Array, etc...

Update

In fact it is in the constructor of List that it figures out if the argument implements ICollection:

public List(IEnumerable<T> collection)
{
  if (collection == null)
    ThrowHelper.ThrowArgumentNullException(ExceptionArgument.collection);
  ICollection<T> collection1 = collection as ICollection<T>;
  if (collection1 != null)
  {
    int count = collection1.Count;
    if (count == 0)
    {
      this._items = List<T>._emptyArray;
    }
    else
    {
      this._items = new T[count];
      collection1.CopyTo(this._items, 0);
      this._size = count;
    }
  }
  else
  {
    this._size = 0;
    this._items = List<T>._emptyArray;
    foreach (T obj in collection)
      this.Add(obj);
  }
}
like image 39
Slugart Avatar answered Oct 17 '22 03:10

Slugart