Using GetHashCode to test equality in Equals override

问题

Is it ok to call GetHashCode as a method to test equality from inside the Equals override?

For example, is this code acceptable?

public class Class1
{
  public string A
  {
    get;
    set;
  }

  public string B
  {
    get;
    set;
  }

  public override bool Equals(object obj)
  {
    Class1 other = obj as Class1;
    return other != null && other.GetHashCode() == this.GetHashCode();
  }

  public override int GetHashCode()
  {
    int result = 0;
    result = (result ^ 397) ^ (A == null ? 0 : A.GetHashCode());
    result = (result ^ 397) ^ (B == null ? 0 : B.GetHashCode());
    return result;
  }
}

回答1:

The others are right; your equality operation is broken. To illustrate:

public static void Main()
{
    var c1 = new Class1() { A = "apahaa", B = null };
    var c2 = new Class1() { A = "abacaz", B = null };
    Console.WriteLine(c1.Equals(c2));
}

I imagine you want the output of that program to be "false" but with your definition of equality it is "true" on some implementations of the CLR.

Remember, there are only about four billion possible hash codes. There are way more than four billion possible six letter strings, and therefore at least two of them have the same hash code. I've shown you two; there are infinitely many more.

In general you can expect that if there are n possible hash codes then the odds of getting a collision rise dramatically once you have about the square root of n elements in play. This is the so-called "birthday paradox". For my article on why you shouldn't rely upon hash codes for equality, see:

http://blogs.msdn.com/b/ericlippert/archive/2010/03/22/socks-birthdays-and-hash-collisions.aspx

回答2:

No, it is not ok, because it's not

equality <=> hashcode equality.

It's just

equality => hashcode equality.

or in the other direction:

hashcode inequality => inequality.

Quoting http://msdn.microsoft.com/en-us/library/system.object.gethashcode.aspx:

If two objects compare as equal, the GetHashCode method for each object must return the same value. However, if two objects do not compare as equal, the GetHashCode methods for the two object do not have to return different values.

回答3:

I would say, unless you want for Equals to basically mean "has the same hash code as" for your type, then no, because two strings may be different but share the same hash code. The probability may be small, but it isn't zero.

回答4:

No this is not an acceptable way to test for equality. It is very possible for 2 non-equal values to have the same hash code. This would cause your implementation of Equals to return true when it should return false

回答5:

You can call GetHashCode to determine if the items are not equal, but if two objects return the same hash code, that doesn't mean they are equal. Two items can have the same hash code but not be equal.

If it's expensive to compare two items, then you can compare the hash codes. If they are unequal, then you can bail. Otherwise (the hash codes are equal), you have to do the full comparison.

For example:

public override bool Equals(object obj)
  {
    Class1 other = obj as Class1;
    if (other == null || other.GetHashCode() != this.GetHashCode())
        return false;
    // the hash codes are the same so you have to do a full object compare.
  }

回答6:

You cannot say that just because the hash codes are equal then the objects must be equal.

The only time you would call GetHashCode inside of Equals was if it was much cheaper to compute a hash value for an object (say, because you cache it) than to check for equality. In that case you could say if (this.GetHashCode() != other.GetHashCode()) return false; so that you could quickly verify that the objects were not equal.

So when would you ever do this? I wrote some code that takes screenshots at periodic intervals and tries to find how long it's been since the screen changed. Since my screenshots are 8MB and have relatively few pixels that change within the screenshot interval it's fairly expensive to search a list of them to find which ones are the same. A hash value is small and only has to be computed once per screenshot, making it easy to eliminate known non-equal ones. In fact, in my application I decided that having identical hashes was close enough to being equal that I didn't even bother to implement the Equals overload, causing the C# compiler to warn me that I was overloading GetHashCode without overloading Equals.

回答7:

There is one case where using hashcodes as a short-cut on equality comparisons makes sense.

Consider the case where you are building a hashtable or hashset. In fact, let's just consider hashsets (hashtables extend that by also holding a value, but that isn't relevant).

There are various different approaches one can take, but in all of them you have a small number of slots the hashed values can be placed in, and we take either the open or closed approach (which just for fun, some people use the opposite jargon for to others); if we collide on the same slot for two different objects we can either store them in the same slot (but having a linked list or such for where the objects are actually stored) or by re-probing to pick a different slot (there are various strategies for this).

Now, with either approach, we're moving away from the O(1) complexity we want with a hashtable, and towards an O(n) complexity. The risk of this is inversely proportional to the number of slots available, so after a certain size we resize the hashtable (even if everything was ideal, we'd eventually have to do this if the number of items stored were greater than the number of slots).

Re-inserting the items on a resize will obviously depend on the hash codes. Because of this, while it rarely makes sense to memoise GetHashCode() in an object (it just isn't called often enough on most objects), it certainly does make sense to memoise it within the hash table itself (or perhaps, to memoise a produced result, such as if you re-hashed with a Wang/Jenkins hash to reduce the damage caused by bad GetHashCode() implementations).

Now, when we come to insert our logic is going to be something like:

Get hash code for object.
Get slot for object.
If slot is empty, place object in it and return.
If slot contains equal object, we're done for a hashset and have the position to replace the value for a hashtable. Do this, and return.
Try next slot according to collision strategy, and return to item 3 (perhaps resizing if we loop this too often).

So, in this case we have to get the hash code before we compare for equality. We also have the hash code for existing objects already pre-computed to allow for resize. The combination of these two facts means that it makes sense to implement our comparison for item 4 as:

private bool IsMatch(KeyType newItem, KeyType storedItem, int newHash, int oldHash)
{
  return ReferenceEquals(newItem, storedItem) // fast, false negatives, no false positives (only applicable to reference types)
    ||
    (
      newHash == oldHash // fast, false positives, no fast negatives
      &&
      _cmp.Equals(newItem, storedItem) // slow for some types, but always correct result.
    );
}

Obviously, the advantage of this depends on the complexity of _cmp.Equals. If our key type was int then this would be a total waste. If our key type where string and we were using case-insensitive Unicode-normalised equality comparisons (so it can't even shortcut on length) then the saving could well be worth it.

Generally memoising hash codes doesn't make sense because they aren't used often enough to be a performance win, but storing them in the hashset or hashtable itself can make sense.

回答8:

It's wrong implementation, as others have stated why.
You should short circuit the equality check using GetHashCode like:
```
if (other.GetHashCode() != this.GetHashCode()
    return false;
```
in the Equals method only if you're certain the ensuing Equals implementation is much more expensive than GetHashCode which is not vast majority of cases.
In this one implementation you have shown (which is 99% of the cases) its not only broken, its also much slower. And the reason? Computing the hash of your properties would almost certainly be slower than comparing them, so you're not even gaining in performance terms. The advantage of implementing a proper GetHashCode is when your class can be the key type for hash tables where hash is computed only once (and that value is used for comparison). In your case GetHashCode will be called multiple times if it's in a collection. Even though GetHashCode itself should be fast, it's not mostly faster than equivalent Equals.

To benchmark, run your Equals (a proper implementation, taking out the current hash based implementation) and GetHashCode here
```
var watch = Stopwatch.StartNew();
for (int i = 0; i < 100000; i++) 
{
    action(); //Equals and GetHashCode called here to test for performance.
}
watch.Stop();
Console.WriteLine(watch.Elapsed.TotalMilliseconds);
```

来源：https://stackoverflow.com/questions/4249064/using-gethashcode-to-test-equality-in-equals-override

标签

gethashcode