What do the return values of Comparable.compareTo

What is the difference between returning 0, returning 1 and returning -1 in compareTo() in Java?

回答1:

Official Definition

From the reference docs of Comparable.compareTo(T):

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

The implementor must ensure sgn(x.compareTo(y)) == -sgn(y.compareTo(x)) for all x and y. (This implies that x.compareTo(y) must throw an exception iff y.compareTo(x) throws an exception.)

The implementor must also ensure that the relation is transitive: (x.compareTo(y)>0 && y.compareTo(z)>0) implies x.compareTo(z)>0.

Finally, the implementor must ensure that x.compareTo(y)==0 implies that sgn(x.compareTo(z)) == sgn(y.compareTo(z)), for all z.

It is strongly recommended, but not strictly required that (x.compareTo(y)==0) == (x.equals(y)). Generally speaking, any class that implements the Comparable interface and violates this condition should clearly indicate this fact. The recommended language is "Note: this class has a natural ordering that is inconsistent with equals."

In the foregoing description, the notation sgn(expression) designates the mathematical signum function, which is defined to return one of -1, 0, or 1 according to whether the value of expression is negative, zero or positive.

My Version

In short:

this.compareTo(that)

returns

a negative int if this < that
0 if this == that
a positive int if this > that

where the implementation of this method determines the actual semantics of < > and == (I don't mean == in the sense of java's object identity operator)

Examples

"abc".compareTo("def")

will yield something smaller than 0 as abc is alphabetically before def.

Integer.valueOf(2).compareTo(Integer.valueOf(1))

will yield something larger than 0 because 2 is larger than 1.

Some additional points

Note: It is good practice for a class that implements Comparable to declare the semantics of it's compareTo() method in the javadocs.

Note: you should read at least one of the following:

the Object Ordering section of the Collection Trail in the Sun Java Tutorial
Effective Java by Joshua Bloch, especially item 12: Consider implementing Comparable
Java Generics and Collections by Maurice Naftalin, Philip Wadler, chapter 3.1: Comparable

Warning: you should never rely on the return values of compareTo being -1, 0 and 1. You should always test for x < 0, x == 0, x > 0, respectively.

回答2:

I use this mnemonic :

a.compareTo(b) < 0 // a < b

a.compareTo(b) > 0 // a > b

a.compareTo(b) == 0 // a == b

You keep the signs and always compare the result of compareTo() to 0

回答3:

Answer in short: (search your situation)

1.compareTo(0) (return: 1)
1.compareTo(1) (return: 0)
0.comapreTo(1) (return: -1)

回答4:

take example if we want to compare "a" and "b", i.e ("a" == this)

negative int if a < b
if a == b
Positive int if a > b

回答5:

It can be used for sorting, and 0 means "equal" while -1, and 1 means "less" and "more (greater)".

Any return value that is less than 0 means that left operand is lesser, and if value is bigger than 0 then left operand is bigger.

回答6:

int x = thisObject.compareTo(anotherObject);

The compareTo() method returns an int with the following characteristics:

negative If thisObject < anotherObject
zero If thisObject == anotherObject
positive If thisObject > anotherObject