Overriding Equals in C# (Part 2)
This post is part two in a series:
(View the completed example)
The Importance of GetHashCode
As soon as you polish off your Equals
implementation, Visual Studio will start whining at you about the fact that you have not overridden a method called GetHashCode
.
So what’s this business about GetHashCode
?
Well, imagine we decide to create a directory that maps phone numbers to employees: for this example, we’ll use our PhoneNumber
class as a key in a Dictionary
:
Dictionary<PhoneNumber, Employee> directory = new Dictionary<PhoneNumber, Employee>();
directory.Add(
new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" },
new Employee { FirstName = "Gordon", LastName = "Freeman" });
directory.Add(
new PhoneNumber { AreaCode = "111", Exchange = "222", SubscriberNumber = "3333" },
new Employee { FirstName = "Samus", LastName = "Aran" });
Great, we have our directory; now let’s try to pull stuff out out of it:
Employee employee = directory[new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" }];
// Throws exception: "The given key was not present in the dictionary"
Whoops, that didn’t work. On top of that, we can do horrifying things like this too:
Dictionary<PhoneNumber, Employee> directory = new Dictionary<PhoneNumber, Employee>();
directory.Add(
new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" },
new Employee { FirstName = "Super", LastName = "Mario" });
directory.Add(
new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" },
new Employee { FirstName = "Princess", LastName = "Peach" });
// No duplicate key exceptions thrown!
So what would work? Well, what if we pass in the original instance:
Dictionary<PhoneNumber, Employee> directory = new Dictionary<PhoneNumber, Employee>();
PhoneNumber number = new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" };
directory.Add(number, new Employee { FirstName = "Super", LastName = "Mario" });
Employee employee = directory[number]; // Works!
As these examples demonstrate, the Dictionary
uses GetHashCode
to quickly pull up a set of results from the collection. Once it has this set, then it can go through each one and verify equality by using Equals
; calling Equals
on a small set retrieved via a hash-based search is a lot more efficient than calling Equals
on the whole lot.
If we haven’t implemented our own version of GetHashCode
, then the default implementation of the method is used, which generates a hash specific to the instance of the object itself, rather than a hash based on the values within our class. To be able to properly use our PhoneNumber
in a hash-based collections, we need the latter.
(You’ll note that this is analogous to how the default, reference equality implementation of Equals
differs from our overridden, value equality implementation above)
Now that we’re convinced we need to override GetHashCode
, let’s do it:
Overriding GetHashCode
Here’s the golden rule for implementing GetHashCode
(from MSDN):
If your overridden Equals method returns true when two objects are tested for equality, your overridden GetHashCode method must return the same value for the two objects.
Based on that, we can surmise that the properties of the class used in our implementation of Equals
should make a reappearance in our implementation of GetHashCode
. On top of that, we know that we need to return an int
from GetHashCode
.
Furthermore, because GetHashCode
will be used to find objects within hash-based collections, our implementation of GetHashCode
should meet two other qualifications: quick to compute with a relatively unique result.
If we have an implementation of GetHashCode
that takes a long time, then we will severely impact the performance of any hash-based collection that uses our class.
Furthermore, recall from above that hash-based collections first use GetHashCode
to get a smaller subset of items from a collection and then look through those objects with Equals
. If our implementation often generates similar hashes, this will cause “collisions” (essentially, similar or identical hashes crowded together) in our hash-based collections, forcing these collections to constantly rifle through a large set of results from a GetHashCode
search.
As a first attempt, we could dust off the exclusive OR operator (^
) and XOR together the individual hash codes of the properties we’re using for equality:
public override int GetHashCode()
{
return AreaCode.GetHashCode() ^ Exchange.GetHashCode() ^ SubscriberNumber.GetHashCode();
}
But this quickly fails on the uniqueness front:
// Two *different* phone numbers
PhoneNumber number1 = new PhoneNumber { AreaCode = "123", Exchange = "456", SubscriberNumber = "7890" };
PhoneNumber number2 = new PhoneNumber { AreaCode = "456", Exchange = "123", SubscriberNumber = "7890" };
number1.GetHashCode() == number2.GetHashCode(); // True
Instead of just adding together our hash numbers, what if we were to multiply our hash numbers by some value (as demonstrated in this StackOverflow post):
public override int GetHashCode()
{
int hash = 13;
hash = (hash * 7) + AreaCode.GetHashCode();
hash = (hash * 7) + Exchange.GetHashCode();
hash = (hash * 7) + SubscriberNumber.GetHashCode();
return hash;
}
In this implementation, we are building an increasingly random product over a series of steps, preventing the collisions we encountered before.
How did we end up with 7 and 13? Well, we were looking for prime numbers as this ensures our hash will produce evenly distributed numbers as explained in this StackOverflow post.
Of course, this is all going to fall apart if one of our properties is null, so we should introduce some null checks:
public override int GetHashCode()
{
int hash = 13;
hash = (hash * 7) + (!Object.ReferenceEquals(null, AreaCode) ? AreaCode.GetHashCode() : 0);
hash = (hash * 7) + (!Object.ReferenceEquals(null, Exchange) ? Exchange.GetHashCode() : 0);
hash = (hash * 7) + (!Object.ReferenceEquals(null, SubscriberNumber) ? SubscriberNumber.GetHashCode() : 0);
return hash;
}
Since we are multiplying together increasingly large integers, it’s likely we will cause an “overflow”. An “overflow” occurs when we reach the maximum value allowed for an int
and then attempt to add more to the int
: the additional amount will “overflow” and we’ll end up starting from the lowest value for int
and working our way backwards.
It makes the most sense when you see an example:
// Int32.MaxValue == 2147483647
int value = 0; // == 0
value += Int32.MaxValue; // == 2147483647
value = value + 10; // == -2147483639
So what? We’re just looking for a random-ish number, right? If .NET wants to quietly wrap our integers around a number line, it’s no big deal to us.
The catch is that you can make this a hard runtime exception: in other words, you can set up your environment to throw an exception in the code sample above (here are instructions on how to do so). In these environments, our GetHashCode
is going to bomb out.
We can prevent this by wrapping our GetHashCode
logic in an unchecked
block (essentially saying, yes, we know overflow will occur and that’s okay):
public override int GetHashCode()
{
unchecked
{
int hash = 13;
hash = (hash * 7) + (!Object.ReferenceEquals(null, AreaCode) ? AreaCode.GetHashCode() : 0);
hash = (hash * 7) + (!Object.ReferenceEquals(null, Exchange) ? Exchange.GetHashCode() : 0);
hash = (hash * 7) + (!Object.ReferenceEquals(null, SubscriberNumber) ? SubscriberNumber.GetHashCode() : 0);
return hash;
}
}
Beautiful! We have ourselves a fine GetHashCode
implementation.
If you read over the comments in response to Jon Skeet’s implementation of GetHashCode, you’ll find two suggested improvements to the hashing algorithm used above:
- Use XOR instead of addition for improved performance (see comment)
- Use a higher prime for the multiplier to further reduce collisions with large data sets
Taking those suggestions, we can implement a modified version of Skeet’s modified version of the FNV hashing algorithm:
public override int GetHashCode()
{
unchecked
{
// Choose large primes to avoid hashing collisions
const int HashingBase = (int) 2166136261;
const int HashingMultiplier = 16777619;
int hash = HashingBase;
hash = (hash * HashingMultiplier) ^ (!Object.ReferenceEquals(null, AreaCode) ? AreaCode.GetHashCode() : 0);
hash = (hash * HashingMultiplier) ^ (!Object.ReferenceEquals(null, Exchange) ? Exchange.GetHashCode() : 0);
hash = (hash * HashingMultiplier) ^ (!Object.ReferenceEquals(null, SubscriberNumber) ? SubscriberNumber.GetHashCode() : 0);
return hash;
}
}
In the next post, we’ll wrap things up by overriding the ==
and !=
operators.