My question may duplicate Default implementation for Object.GetHashCode() but I'm asking again because I didn't understand the accepted answer to that one.
To begin with I have three questions about the accepted answer to the previous question, which quotes some documentation as follows:
"However, because this index can be reused after the object is reclaimed during garbage collection, it is possible to obtain the same hash code for two different objects."
Is this true? It seems to me that two objects won't have the same hash code, because an object's code isn't reused until the object is garbage collected (i.e. no longer exists).
"Also, two objects that represent the same value have the same hash code only if they are the exact same object."
Is this a problem? For example, I want to associate some data with each of the node instances in a DOM tree. To do this, the 'nodes' must have an identity or hash code, so that I can use them as keys in a dictionary of the data. Isn't a hash code which identities whether it's "the exact same object", i.e. "reference equality rather than "value equality", what I want?
"This implementation is not particularly useful for hashing; therefore, derived classes should override GetHashCode"
Is this true? If it's not good for hashing, then what if anything is it good for, and why is it even defined as a method of Object?
My final (and perhaps most important to me) question is, if I must invent/override a GetHashCode() implementation for an arbitrary type which has "reference equality" semantics, is the following a reasonable and good implementation:
class SomeType
{
//create a new value for each instance
static int s_allocated = 0;
//value associated with this instance
int m_allocated;
//more instance data
... plus other data members ...
//constructor
SomeType()
{
allocated = ++s_allocated;
}
//override GetHashCode
public override int GetHashCode()
{
return m_allocated;
}
}
Edit
FYI I tested it, using the following code:
class TestGetHash
{
//default implementation
class First
{
int m_x;
}
//my implementation
class Second
{
static int s_allocated = 0;
int m_allocated;
int m_x;
public Second()
{
m_allocated = ++s_allocated;
}
public override int GetHashCode()
{
return m_allocated;
}
}
//stupid worst-case implementation
class Third
{
int m_x;
public override int GetHashCode()
{
return 0;
}
}
internal static void test()
{
testT<First>(100, 1000);
testT<First>(1000, 100);
testT<Second>(100, 1000);
testT<Second>(1000, 100);
testT<Third>(100, 100);
testT<Third>(1000, 10);
}
static void testT<T>(int objects, int iterations)
where T : new()
{
System.Diagnostics.Stopwatch stopWatch =
System.Diagnostics.Stopwatch.StartNew();
for (int i = 0; i < iterations; ++i)
{
Dictionary<T, object> dictionary = new Dictionary<T, object>();
for (int j = 0; j < objects; ++j)
{
T t = new T();
dictionary.Add(t, null);
}
for (int k = 0; k < 100; ++k)
{
foreach (T t in dictionary.Keys)
{
object o = dictionary[t];
}
}
}
stopWatch.Stop();
string stopwatchMessage = string.Format(
"Stopwatch: {0} type, {1} objects, {2} iterations, {3} msec",
typeof(T).Name, objects, iterations,
stopWatch.ElapsedMilliseconds);
System.Console.WriteLine(stopwatchMessage);
}
}
On my machine the results/output are as follows:
First type, 100 objects, 1000 iterations, 2072 msec
First type, 1000 objects, 100 iterations, 2098 msec
Second type, 100 objects, 1000 iterations, 1300 msec
Second type, 1000 objects, 100 iterations, 1319 msec
Third type, 100 objects, 100 iterations, 1487 msec
Third type, 1000 objects, 10 iterations, 13754 msec
My implementation takes half the time of the default implementation (but my type is bigger by the size of my m_allocated data member).
My implementation and the default implementation both scale linearly.
In comparison and as a sanity check, the stupid implementation starts bad and scales worse.
Answer
The most important property a hash code implementation must have is this:
If two objects compare as equal then they must have identical hash codes.
If you have a class where instances of the class compare by reference equality, then you do not need to override GetHashCode; the default implementation guarantees that two objects that are the same reference have the same hash code. (You're calling the same method twice on the same object, so of course the result is the same.)
If you have written a class which implements its own equality that is different from reference equality then you are REQUIRED to override GetHashCode such that two objects that compare as equal have equal hash codes.
Now, you could do so by simply returning zero every time. That would be a lousy hash function, but it would be legal.
Other properties of good hash functions are:
GetHashCode should never throw an exception
Mutable objects which compare for equality on their mutable state, and therefore hash on their mutable state, are dangerously bug-prone. You can put an object into a hash table, mutate it, and be unable to get it out again. Try to never hash or compare for equality on mutable state.
GetHashCode should be extremely fast -- remember, the purpose of a good hash algorithm is to improve the performance of lookups. If the hash is slow then the lookups can't be made fast.
Objects which do not compare as equal should have dissimilar hash codes, well distributed over the whole range of a 32 bit integer