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Comparing two values for equality is ubiquitous in programming. It is also more tricky than it looks at first glance. This chapter looks at object equality in detail and gives some recommendations to consider when you design your own equality tests.
As mentioned in Section 11.2, the definition of equality is different in Scala and Java. Java has two equality comparisons: the == operator, which is the natural equality for value types and object identity for reference types, and the equals method, which is (user-defined) canonical equality for reference types. This convention is problematic, because the more natural symbol, ==, does not always correspond to the natural notion of equality. When programming in Java, a common pitfall for beginners is to compare objects with == when they should have been compared with equals. For instance, comparing two strings x and y using "x == y" might well yield false in Java, even if x and y have exactly the same characters in the same order.
Scala also has an equality method signifying object identity, but it is not used much. That kind of equality, written "x eq y", is true if x and y reference the same object. The == equality is reserved in Scala for the "natural" equality of each type. For value types, == is value comparison, just like in Java. For reference types, == is the same as equals in Scala. You can redefine the behavior of == for new types by overriding the equals method, which is always inherited from class Any. The inherited equals, which takes effect unless overridden, is object identity, as is the case in Java. So equals (and with it, ==) is by default the same as eq, but you can change its behavior by overriding the equals method in the classes you define. It is not possible to override == directly, as it is defined as a final method in class Any. That is, Scala treats == as if it were defined as follows in class Any:
final def == (that: Any): Boolean = if (null eq this) {null eq that} else {this equals that}
How should the equals method be defined? It turns out that writing a correct equality method is surprisingly difficult in object-oriented languages. In fact, after studying a large body of Java code, the authors of a 2007 paper concluded that almost all implementations of equals methods are faulty.[1]
This is problematic, because equality is at the basis of many other things. For one, a faulty equality method for a type C might mean that you cannot reliably put an object of type C in a collection. You might have two elements elem1, elem2 of type C which are equal, i.e., "elem1 equals elem2" yields true. Nevertheless, with commonly occurring faulty implementations of the equals method, you could still see behavior like the following:
var hashSet: Set[C] = new collection.immutable.HashSet hashSet += elem1 hashSet contains elem2 // returns false!Here are four common pitfalls[2] that can cause inconsistent behavior when overriding equals:
Consider adding an equality method to the following class of simple points:
class Point(val x: Int, val y: Int) { ... }A seemingly obvious, but wrong way would be to define it like this:
// An utterly wrong definition of equals def equals(other: Point): Boolean = this.x == other.x && this.y == other.yWhat's wrong with this method? At first glance, it seems to work OK:
scala> val p1, p2 = new Point(1, 2) p1: Point = Point@62d74e p2: Point = Point@254de0However, trouble starts once you start putting points into a collection:
scala> val q = new Point(2, 3) q: Point = Point@349f8a
scala> p1 equals p2 res0: Boolean = true
scala> p1 equals q res1: Boolean = false
scala> import scala.collection.mutable._ import scala.collection.mutable._How to explain that coll does not contain p2, even though p1 was added to it, and p1 and p2 are equal objects? The reason becomes clear in the following interaction, where the precise type of one of the compared points is masked. Define p2a as an alias of p2, but with type Any instead of Point:
scala> val coll = HashSet(p1) coll: scala.collection.mutable.Set[Point] = Set(Point@62d74e)
scala> coll contains p2 res2: Boolean = false
scala> val p2a: Any = p2 p2a: Any = Point@254de0Now, were you to repeat the first comparison, but with the alias p2a instead of p2, you would get:
scala> p1 equals p2a
res3: Boolean = false
What went wrong? In fact, the version of equals given previously does not override
the standard method equals, because its type is different. Here is the
type of the equals method as it is defined in the root class
Any:[3]
def equals(other: Any): BooleanBecause the equals method in Point takes a Point instead of an Any as an argument, it does not override equals in Any. Instead, it is just an overloaded alternative. Now, overloading in Scala and in Java is resolved by the static type of the argument, not the run-time type. So as long as the static type of the argument is Point, the equals method in Point is called. However, once the static argument is of type Any, the equals method in Any is called instead. This method has not been overridden, so it is still implemented by comparing object identity. That's why the comparison "p1 equals p2a" yields false even though points p1 and p2a have the same x and y values. That's also why the contains method in HashSet returned false. Since that method operates on generic sets, it calls the generic equals method in Object instead of the overloaded variant in Point.
A better equals method is the following:
// A better definition, but still not perfect override def equals(other: Any) = other match { case that: Point => this.x == that.x && this.y == that.y case _ => false }Now equals has the correct type. It takes a value of type Any as parameter and it yields a Boolean result. The implementation of this method uses a pattern match. It first tests whether the other object is also of type Point. If it is, it compares the coordinates of the two points and returns the result. Otherwise the result is false.
A related pitfall is to define == with a wrong signature. Normally, if you try to redefine == with the correct signature, which takes an argument of type Any, the compiler will give you an error because you try to override a final method of type Any. However, newcomers to Scala sometimes make two errors at once: They try to override == and they give it the wrong signature. For instance:
def ==(other: Point): Boolean = // Don't do this!In that case, the user-defined == method is treated as an overloaded variant of the same-named method class Any, and the program compiles. However, the behavior of the program would be just as dubious as if you had defined equals with the wrong signature.
If you repeat the comparison of p1 and p2a with the latest definition of Point defined previously, you will get true, as expected. However, if you repeat the HashSet.contains test, you will probably still get false.
scala> val p1, p2 = new Point(1, 2) p1: Point = Point@670f2b p2: Point = Point@14f7c0In fact, this outcome is not 100% certain. You might also get true from the experiment. If you do, you can try with some other points with coordinates 1 and 2. Eventually, you'll get one which is not contained in the set. What goes wrong here is that Point redefined equals without also redefining hashCode.
scala> HashSet(p1) contains p2 res4: Boolean = false
Note that the collection in the example above is a HashSet. This means elements of the collection are put in "hash buckets" determined by their hash code. The contains test first determines a hash bucket to look in and then compares the given elements with all elements in that bucket. Now, the last version of class Point did redefine equals, but it did not at the same time redefine hashCode. So hashCode is still what it was in its version in class AnyRef: some transformation of the address of the allocated object. The hash codes of p1 and p2 are almost certainly different, even though the fields of both points are the same. Different hash codes mean with high probability different hash buckets in the set. The contains test will look for a matching element in the bucket which corresponds to p2's hash code. In most cases, point p1 will be in another bucket, so it will never be found. p1 and p2 might also end up by chance in the same hash bucket. In that case the test would return true.
The problem was that the last implementation of Point violated the contract on hashCode as defined for class Any:[4]
If two objects are equal according to the equals method, then calling the hashCode method on each of the two objects must produce the same integer result.
In fact, it's well known in Java that hashCode and equals should always be redefined together. Furthermore, hashCode may only depend on fields that equals depends on. For the Point class, the following would be a suitable definition of hashCode:
class Point(val x: Int, val y: Int) { override def hashCode = 41 * (41 + x) + y override def equals(other: Any) = other match { case that: Point => this.x == that.x && this.y == that.y case _ => false } }This is just one of many possible implementations of hashCode. Adding the constant 41 to one integer field x, multiplying the result with the prime number 41, and adding to that result the other integer field y gives a reasonable distribution of hash codes at a low cost in running time and code size. We'll provide more guidance on writing hashCode later in this chapter.
Adding hashCode fixes the problems of equality when defining classes like Point. However, there are still other trouble spots to watch out for.
Consider the following slight variation of class Point:
class Point(var x: Int, var y: Int) { // Problematic override def hashCode = 41 * (41 + x) + y override def equals(other: Any) = other match { case that: Point => this.x == that.x && this.y == that.y case _ => false } }The only difference is that the fields x and y are now vars instead of vals. The equals and hashCode methods are now defined in terms of these mutable fields, so their results change when the fields change. This can have strange effects once you put points in collections:
scala> val p = new Point(1, 2) p: Point = Point@2bNow, if you change a field in point p, does the collection still contain the point? We'll try it:
scala> val coll = HashSet(p) coll: scala.collection.mutable.Set[Point] = Set(Point@2b)
scala> coll contains p res5: Boolean = true
scala> p.x += 1This looks strange. Where did p go? More strangeness results if you check whether the elements iterator of the set contains p:
scala> coll contains p res7: Boolean = false
scala> coll.elements contains p
res8: Boolean = true
So here's a set that does not contain p, yet p is among the elements of the set!
What happened, of course, is that after the change to the x field, the point p ended
up in the wrong hash bucket of the set coll. That is, its original hash bucket
no longer corresponded to the new value of its hash code.
In a manner of speaking, the point p "dropped out of sight"
in the set coll even though it still belonged to its elements.
The lesson to be drawn from this example is that when equals and hashCode depend on mutable state, it causes problems for potential users. If they put such objects into collections, they have to be careful never to modify the depended-on state, and this is tricky. If you need a comparison that takes the current state of an object into account, you should usually name it something else, not equals. Considering the last definition of Point, it would have been preferable to omit a redefinition of hashCode and to name the comparison method equalContents, or some other name different from equals. Point would then have inherited the default implementation of equals and hashCode. So p would have stayed locatable in coll even after the modification to its x field.
The contract of the equals method in scala.Any specifies that equals must implement an equivalence relation on non-null objects:[5]
The definition of equals developed so far for class Point satisfies the contract for equals. However, things become more complicated once subclasses are considered. Say there is a subclass ColoredPoint of Point that adds a field color of type Color. Assume Color is defined as an enumeration, as presented in Section 20.8:
object Color extends Enumeration { val Red, Orange, Yellow, Green, Blue, Indigo, Violet = Value }ColoredPoint overrides equals to take the new color field into account:
class ColoredPoint(x: Int, y: Int, val color: Color.Value) extends Point(x, y) { // Problem: equals not symmetricThis is what many programmers would likely write. Note that in this case, class ColoredPoint need not override hashCode. Because the new definition of equals on ColoredPoint is stricter than the overridden definition in Point (meaning it equates fewer pairs of objects), the contract for hashCode stays valid. If two colored points are equal, they must have the same coordinates, so their hash codes are guaranteed to be equal as well.
override def equals(other: Any) = other match { case that: ColoredPoint => this.color == that.color && super.equals(that) case _ => false } }
Taking the class ColoredPoint by itself, its definition of equals looks OK. However, the contract for equals is broken once points and colored points are mixed. Consider:
scala> val p = new Point(1, 2) p: Point = Point@2bThe comparison "p equals cp" invokes p's equals method, which is defined in class Point. This method only takes into account the coordinates of the two points. Consequently, the comparison yields true. On the other hand, the comparison "cp equals p" invokes cp's equals method, which is defined in class ColoredPoint. This method returns false, because p is not a ColoredPoint. So the relation defined by equals is not symmetric.
scala> val cp = new ColoredPoint(1, 2, Color.Red) cp: ColoredPoint = ColoredPoint@2b
scala> p equals cp res8: Boolean = true
scala> cp equals p res9: Boolean = false
The loss in symmetry can have unexpected consequences for collections. Here's an example:
scala> HashSet[Point](p) contains cp res10: Boolean = trueSo even though p and cp are equal, one contains test succeeds whereas the other one fails.
scala> HashSet[Point](cp) contains p res11: Boolean = false
How can you change the definition of equals so that it becomes symmetric? Essentially there are two ways. You can either make the relation more general or more strict. Making it more general means that a pair of two objects, x and y, is taken to be equal if either comparing x with y or comparing y with x yields true. Here's code that does this:
class ColoredPoint(x: Int, y: Int, val color: Color.Value) extends Point(x, y) { // Problem: equals not transitiveThe new definition of equals in ColoredPoint has one more case than the old one: If the other object is a Point but not a ColoredPoint, the method forwards to the equals method of Point. This has the desired effect of making equals symmetric. Now, both "cp equals p" and "p equals cp" result in true. However, the contract for equals is still broken. Now the problem is that the new relation is no longer transitive! Here's a sequence of statements that demonstrates this. Define a point and two colored points of different colors, all at the same position:
override def equals(other: Any) = other match { case that: ColoredPoint => (this.color == that.color) && super.equals(that) case that: Point => that equals this case _ => false } }
scala> val redp = new ColoredPoint(1, 2, Color.Red) redp: ColoredPoint = ColoredPoint@2bTaken individually, redp is equal to p and p is equal to bluep:
scala> val bluep = new ColoredPoint(1, 2, Color.Blue) bluep: ColoredPoint = ColoredPoint@2b
scala> redp == p res12: Boolean = trueHowever, comparing redp and bluep yields false:
scala> p == bluep res13: Boolean = true
scala> redp == bluep
res14: Boolean = false
Hence, the transitivity clause of equals's contract is violated.
Making the equals relation more general seems to lead to a dead end.
We'll try to make it stricter instead. One way to make equals stricter is
to always treat objects of different classes as different. That could be achieved by
modifying the equals methods in classes Point and ColoredPoint.
In class Point, you
could add an extra comparison that checks whether the run-time class of the other Point is exactly the same
as this Point's class, as follows:
// A technically valid, but unsatisfying, equals method class Point(val x: Int, val y: Int) { override def hashCode = 41 * (41 + x) + y override def equals(other: Any) = other match { case that: Point => this.x == that.x && this.y == that.y && this.getClass == that.getClass case _ => false } }You can then revert class ColoredPoint's implementation back to the version that previously had violated the symmetry requirement:[6]
class ColoredPoint(x: Int, y: Int, val color: Color.Value) extends Point(x, y) {Here, an instance of class Point is considered to be equal to some other instance of the same class only if the objects have the same coordinates and they have the same run-time class, meaning .getClass on either object returns the same value. The new definitions satisfy symmetry and transitivity because now every comparison between objects of different classes yields false. So a colored point can never be equal to a point. This convention looks reasonable, but one could argue that the new definition is too strict.
override def equals(other: Any) = other match { case that: ColoredPoint => (this.color == that.color) && super.equals(that) case _ => false } }
Consider the following slightly roundabout way to define a point at coordinates (1, 2):
scala> val pAnon = new Point(1, 1) { override val y = 2 } pAnon: Point = $anon$1@2bIs pAnon equal to p? The answer is no because the java.lang.Class objects associated with p and pAnon are different. For p it is Point, whereas for pAnon it is an anonymous class of Point. But clearly, pAnon is just another point at coordinates (1, 2). It does not seem reasonable to treat it as being different from p.
So it seems we are stuck. Is there a sane way to redefine equality on several levels of the class hierarchy while keeping its contract? In fact, there is such a way, but it requires one more method to redefine together with equals and hashCode. The idea is that as soon as a class redefines equals (and hashCode), it should also explicitly state that objects of this class are never equal to objects of some superclass that implement a different equality method. This is achieved by adding a method canEqual to every class that redefines equals. Here's the method's signature:
def canEqual(other: Any): BooleanThe method should return true if the other object is an instance of the class in which canEqual is (re)defined, false otherwise. It is called from equals to make sure that the objects are comparable both ways. Listing 28.1 shows a new (and final) implementation of class Point along these lines:
class Point(val x: Int, val y: Int) { override def hashCode = 41 * (41 + x) + y override def equals(other: Any) = other match { case that: Point => (that canEqual this) && (this.x == that.x) && (this.y == that.y) case _ => false } def canEqual(other: Any) = other.isInstanceOf[Point] }
The equals method in this version of class Point contains the additional requirement that the other object can equal this one, as determined by the canEqual method. The implementation of canEqual in Point states that all instances of Point can be equal.
Listing 28.2 shows the corresponding implementation of ColoredPoint.
class ColoredPoint(x: Int, y: Int, val color: Color.Value) extends Point(x, y) {
override def hashCode = 41 * super.hashCode + color.hashCode override def equals(other: Any) = other match { case that: ColoredPoint => (that canEqual this) && super.equals(that) && this.color == that.color case _ => false } override def canEqual(other: Any) = other.isInstanceOf[ColoredPoint] }
It can be shown that the new definition of Point and ColoredPoint keeps the contract of equals. Equality is symmetric and transitive. Comparing a Point to a ColoredPoint always yields false. Indeed, for any point p and colored point cp, "p equals cp" will return false because "cp canEqual p" will return false. The reverse comparison, "cp equals p", will also return false, because p is not a ColoredPoint, so the first pattern match in the body of equals in ColoredPoint will fail.
On the other hand, instances of different subclasses of Point can be equal, as long as none of the classes redefines the equality method. For instance, with the new class definitions, the comparison of p and pAnon would yield true. Here are some examples:
scala> val p = new Point(1, 2) p: Point = Point@6bcThese examples demonstrate that if a superclass equals implementation defines and calls canEqual, then programmers who implement subclasses can decide whether or not their subclasses may be equal to instances of the superclass. Because ColoredPoint overrides canEqual, for example, a colored point may never be equal to a plain-old point. But because the anonymous subclass referenced from pAnon does not override canEqual, its instance can be equal to a Point instance.
scala> val cp = new ColoredPoint(1, 2, Color.Indigo) cp: ColoredPoint = ColoredPoint@11421
scala> val pAnon = new Point(1, 1) { override val y = 2 } pAnon: Point = $anon$1@6bc
scala> val coll = List(p) coll: List[Point] = List(Point@6bc)
scala> coll contains p res0: Boolean = true
scala> coll contains cp res1: Boolean = false
scala> coll contains pAnon res2: Boolean = true
One potential criticism of the canEqual approach is that it violates the Liskov Substitution Principle (LSP). For example, the technique of implementing equals by comparing run-time classes, which led to the inability to define a subclass whose instances can equal instances of the superclass, has been described as a violation of the LSP.[7] The reasoning is that the LSP states you should be able to use (substitute) a subclass instance where a superclass instance is required. In the previous example, however, "coll contains cp" returned false even though cp's x and y values matched those of the point in the collection. Thus it may seem like a violation of the LSP, because you can't use a ColoredPoint here where a Point is expected. We believe this is the wrong interpretation, though, because the LSP doesn't require that a subclass behaves identically to its superclass, just that it behaves in a way that fulfills the contract of its superclass.
The problem with writing an equals method that compares run-time classes is not that it violates the LSP, but that it doesn't give you a way to create a subclass whose instances can equal superclass instances. For example, had we used the run-time class technique in the previous example, "coll contains pAnon" would have returned false, and that's not what we wanted. By contrast, we really did want "coll contains cp" to return false, because by overriding equals in ColoredPoint, we were basically saying that an indigo-colored point at coordinates (1, 2) is not the same thing as an uncolored point at (1, 2). Thus, in the previous example we were able to pass two different Point subclass instances to the collection's contains method, and we got back two different answers, both correct.
The equals methods in the previous examples all started with a pattern match that tested whether the type of the operand conformed to the type of the class containing the equals method. When classes are parameterized, this scheme needs to be adapted a little bit. As an example, consider binary trees. The class hierarchy shown in Listing 28.3 defines an abstract class Tree for a binary tree, with two alternative implementations: an EmptyTree object and a Branch class representing non-empty trees. A non-empty tree is made up of some element elem and a left and right child tree. The type of its element is given by a type parameter T.
trait Tree[+T] { def elem: T def left: Tree[T] def right: Tree[T] }
object EmptyTree extends Tree[Nothing] { def elem = throw new NoSuchElementException("EmptyTree.elem") def left = throw new NoSuchElementException("EmptyTree.left") def right = throw new NoSuchElementException("EmptyTree.right") }
class Branch[+T]( val elem: T, val left: Tree[T], val right: Tree[T] ) extends Tree[T]
We'll now add equals and hashCode methods to these classes. For class Tree itself there's nothing to do, because we assume that these methods are implemented separately for each implementation of the abstract class. For object EmptyTree, there's still nothing to do because the default implementations of equals and hashCode that EmptyTree inherits from AnyRef work just fine. After all, an EmptyTree is only equal to itself, so equality should be reference equality, which is what's inherited from AnyRef.
But adding equals and hashCode to Branch requires some work. A Branch value should only be equal to other Branch values, and only if the two values have equal elem, left and right fields. It's natural to apply the schema for equals that was developed in the previous sections of this chapter. This would give:
class Branch[T]( val elem: T, val left: Tree[T], val right: Tree[T] ) extends Tree[T] {Compiling this example, however, gives an indication that "unchecked" warnings occurred. Compiling again with the -unchecked option reveals the following problem:
override def equals(other: Any) = other match { case that: Branch[T] => this.elem == that.elem && this.left == that.left && this.right == that.right case _ => false } }
$ fsc -unchecked Tree.scala Tree.scala:14: warning: non variable type-argument T in type pattern is unchecked since it is eliminated by erasure case that: Branch[T] => this.elem == that.elem && ^
As the warning says, there is a pattern match against a Branch[T] type, yet the system can only check that the other reference is (some kind of) Branch; it cannot check that the element type of the tree is T. You encountered in Chapter 19 the reason for this: element types of parameterized types are eliminated by the compiler's erasure phase; they are not available to be inspected at run-time.
So what can you do? Fortunately, it turns out that you need not necessarily check that two Branches have the same element types when comparing them. It's quite possible that two Branches with different element types are equal, as long as their fields are the same. A simple example of this would be the Branch that consists of a single Nil element and two empty subtrees. It's plausible to consider any two such Branches to be equal, no matter what static types they have:
scala> val b1 = new Branch[List[String]](Nil, EmptyTree, EmptyTree) b1: Branch[List[String]] = Branch@2f1eb9
scala> val b2 = new Branch[List[Int]](Nil, EmptyTree, EmptyTree) b2: Branch[List[Int]] = Branch@be55d1
scala> b1 == b2 res0: Boolean = true
The positive result of the comparison above was obtained with the implementation of equals on Branch shown previously. This demonstrates that the element type of the Branch was not checked—if it had been checked, the result would have been false.
Note that one can disagree which of the two possible outcomes of the comparison would be more natural. In the end, this depends on the mental model of how classes are represented. In a model where type-parameters are present only at compile-time, it's natural to consider the two Branch values b1 and b2 to be equal. In an alternative model where a type parameter forms part of an object's value, it's equally natural to consider them different. Since Scala adopts the type erasure model, type parameters are not preserved at run time, so that b1 and b2 are naturally considered to be equal.
There's only a tiny change needed to formulate an equals method that does not produce an unchecked warning: instead of an element type T, use a lower case letter, such as t:
case that: Branch[t] => this.elem == that.elem && this.left == that.left && this.right == that.rightRecall from Section 15.2 that a type parameter in a pattern starting with a lower-case letter represents an unknown type. Hence, the pattern match:
case that: Branch[t] =>will succeed for Branch values of any type. The type parameter t represents the unknown element type of the Branch. It can also be replaced by an underscore, as in the following case, which is equivalent to the previous one:
case that: Branch[_] =>
The only thing that remains is to define for class Branch the other two methods, hashCode and canEqual, which go with equals. Here's a possible implementation of hashCode:
override def hashCode: Int = 41 * ( 41 * ( 41 + elem.hashCode ) + left.hashCode ) + right.hashCodeThis is only one of many possible implementations. As shown previously, the principle is to take hashCode values of all fields, and to combine them using additions and multiplications by some prime number. Here's an implementation of method canEqual in class Branch:
def canEqual(other: Any) = other match { case that: Branch[_] => true case _ => false }The implementation of the canEqual method used a typed pattern match. It would also be possible to formulate it with isInstanceOf:
def canEqual(other: Any) = other.isInstanceOf[Branch[_]]If you feel like nit-picking (and we encourage you to do so!), you might wonder what the occurrence of the underscore in the type above signifies. After all, Branch[_] is a technically a type parameter of a method, not a type pattern, so how is it possible to leave some parts of it undefined? The answer to that question is found in the next chapter: Branch[_] is a shorthand for a so-called existential type, which is roughly speaking a type with some unknown parts in it. So even though technically the underscore stands for two different things in a pattern match and in a type parameter of a method call, in essence the meaning is the same: it lets you label something that is unknown. The final version of Branch is shown in Listing 28.4.
class Branch[T]( val elem: T, val left: Tree[T], val right: Tree[T] ) extends Tree[T] {
override def equals(other: Any) = other match { case that: Branch[_] => (that canEqual this) && this.elem == that.elem && this.left == that.left && this.right == that.right case _ => false }
def canEqual(other: Any) = other.isInstanceOf[Branch[_]]
override def hashCode: Int = 41 * ( 41 * ( 41 + elem.hashCode ) + left.hashCode ) + right.hashCode }
In this section, we'll provide step-by-step recipes for creating equals and hashCode methods that should suffice for most situations. As an illustration, we'll use the methods of class Rational, shown in Listing 28.5. To create this class, we removed the mathematical operator methods from the version of class Rational shown in Listing 6.5 here. We also made a minor enhancement to toString and modified the initializers of numer and denom to normalize all fractions to have a positive denominator (i.e., to transform 1/-2 to -1/2). Here's the recipe for overriding equals:
class Rational(n: Int, d: Int) {
require(d != 0)
private val g = gcd(n.abs, d.abs) val numer = (if (d < 0) -n else n) / g val denom = d.abs / g
private def gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
override def equals(other: Any): Boolean = other match {
case that: Rational => (that canEqual this) && numer == that.numer && denom == that.denom
case _ => false }
def canEqual(other: Any): Boolean = other.isInstanceOf[Rational]
override def hashCode: Int = 41 * ( 41 + numer ) + denom
override def toString = if (denom == 1) numer.toString else numer +"/"+ denom }
def canEqual(other: Any): Boolean =
other.isInstanceOf[Rational]
override def equals(other: Any): Boolean =
other match { // ... }
case that: Rational =>
super.equals(that) &&
If you are defining equals for a class that first introduced canEqual, you should invoke canEqual on the
argument to the equality method, passing this as the argument:
(that canEqual this) &&
Overriding redefinitions of equals should also include the canEqual invocation, unless they contain
a call to super.equals. In the latter case, the canEqual test will already be
done by the superclass call.
Lastly, for each field relevant to equality, verify that the field in this object is equal to the corresponding field in the passed object:
numer == that.numer && denom == that.denom
case _ => false
If you adhere to the preceding recipe, equality is guaranteed to be an equivalence relation, as is required by the equals contract.
For hashCode, you can usually achieve satisfactory results if you use the following recipe, which is similar to a recipe recommended for Java classes in Effective Java.[8] Include in the calculation each field in your object that is used to determine equality in the equals method (the "relevant" fields). For each relevant field, no matter its type, you can calculate a hash code by invoking hashCode on it. To calculate a hash code for the entire object, add 41 to the first field's hash code, multiply that by 41, add the second field's hash code, multiply that by 41, add the third field's hash code, multiply that by 41, and so on, until you've done this for all relevant fields.
For example, to implement the hash code for an object that has five relevant fields named a, b, c, d, and e, you would write:
override def hashCode: Int = 41 * ( 41 * ( 41 * ( 41 * ( 41 + a.hashCode ) + b.hashCode ) + c.hashCode ) + d.hashCode ) + e.hashCodeIf you wish, you can leave off the hashCode invocation on fields of type Int, Short, Byte, and Char. The hash code for an Int is the value of the Int, as are the hash codes of Shorts, Bytes, and Chars when automatically widened to Int. Given numer or denom are Ints, therefore, we implemented Rational's hashCode method like this:
override def hashCode: Int = 41 * ( 41 + numer ) + denomThe number 41 was selected for the multiplications because it is an odd prime. You could use another number, but it should be an odd prime to minimize the potential for information loss on overflow. The reason we added 41 to the innermost value is to reduce the likelihood that the first multiplication will result in zero, under the assumption that it is more likely the first field used will be zero than -41. The number 41 was chosen for the addition only for looks. You could use any non-zero integer.
If the equals method invokes super.equals(that) as part of its calculation, you should start your hashCode calculation with an invocation of super.hashCode. For example, had Rational's equals method invoked super.equals(that), its hashCode would have been:
override def hashCode: Int = 41 * ( 41 * ( super.hashCode ) + numer ) + denom
One thing to keep in mind as you write hashCode methods using this approach is that your hash code will only be as good as the hash codes you build it out of, namely the hash codes you obtain by calling hashCode on the relevant fields of your object. Sometimes you may need to do something extra besides just calling hashCode on the field to get a useful hash code for that field. For example, if one of your fields is a collection, you probably want a hash code for that field that is based on all the elements contained in the collection. If the field is a List, Set, Map, or tuple, you can simply call hashCode on the field, because equals and hashCode are overridden in those classes to take into account the contained elements. However the same is not true for Arrays, which do not take elements into account when calculating a hash code. Thus for an array, you should treat each element of the array like an individual field of your object, calling hashCode on each element explicitly, or passing the array to one of the hashCode methods in singleton object java.util.Arrays.
Lastly, if you find that a particular hash code calculation is harming the performance of your program, you can consider caching the hash code. If the object is immutable, you can calculate the hash code when the object is created and store it in a field. You can do this by simply overriding hashCode with a val instead of a def, like this:
override val hashCode: Int = 41 * ( 41 + numer ) + denomThis approach trades off memory for computation time, because each instance of the immutable class will have one more field to hold the cached hash code value.
In retrospect, defining a correct implementation of equals has been surprisingly subtle. You must be careful about the type signature; you must override hashCode; you should avoid dependencies on mutable state; and you should implement and use a canEqual method if your class is non-final. Given how difficult it is to implement a correct equality method, you might prefer to define your classes of comparable objects as case classes. That way, the Scala compiler will add equals and hashCode methods with the right properties automatically.
[1] Vaziri, et al., "Declarative Object Identity Using Relation Types" vaziri-tip-2007
[2] All but the third of these pitfalls are described in the context of Java in the book, Effective Java Second Edition, by Joshua Bloch. bloch:effective-java
[3] If you write a lot of Java, you might expect the argument to this method to be type Object instead of type Any. Don't worry about it. It is the same equals method. The compiler simply makes it appear to have type Any.
[4] The text of Any's hashCode contract is inspired by the Javadoc documentation of class java.lang.Object.
[5] As with hashCode, Any's equals contract is based on the contract of equals in java.lang.Object.
[6] Given the new implementation of equals in Point, this version of ColoredPoint no longer violates the symmetry requirement.
[7] Bloch, Effective Java Second Edition, p. 39 bloch:effective-java
[8] Bloch, Effective Java Second Edition. bloch:effective-java