AbstractFunction2

f(x1, x2) == (f.curried)(x1)(x2)

This method is used to compare the receiver object (this) with the argument object (arg0) for equivalence.

The default implementations of this method is an equivalence relation:

• It is reflexive: for any instance x of type Any, x.equals(x) should return true.
• It is symmetric: for any instances x and y of type Any, x.equals(y) should return true if and only if y.equals(x) returns true.
• It is transitive: for any instances x, y, and z of type AnyRef if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.

If you override this method, you should verify that your implementation remains an equivalence relation. Additionally, when overriding this method it is often necessary to override hashCode to ensure that objects that are "equal" (o1.equals(o2) returns true) hash to the same Int (o1.hashCode.equals(o2.hashCode)).

Returns a hash code value for the object.

The default hashing algorithm is platform dependent.

Note that it is allowed for two objects to have identical hash codes (o1.hashCode.equals(o2.hashCode)) yet not be equal (o1.equals(o2) returns false). A degenerate implementation could always return 0. However, it is required that if two objects are equal (o1.equals(o2) returns true) that they have identical hash codes (o1.hashCode.equals(o2.hashCode)). Therefore, when overriding this method, be sure to verify that the behavior is consistent with the equals method.

Returns a string representation of the object.

The default representation is platform dependent.