java - compareTo() 实际上返回什么？

Question

Java 中的 compareTo() 方法返回一个大于/等于/小于 0 的值，我知道这一点。但是，值(value)本身就是我的问题。 compareTo() 返回时 2 或 4 之间有什么区别。看下面的代码

String s1="hello";  
String s2="hello";  
String s3="meklo";  
String s4="hemlo";  
System.out.println(s1.compareTo(s2));     // 0
System.out.println(s1.compareTo(s3));     // -5
System.out.println(s1.compareTo(s4));     // -1

为什么最后两个命令是-5和-1？

Answer 1

https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#compareTo(java.lang.String)

This is the definition of lexicographic ordering. If two strings are different, then either they have different characters at some index that is a valid index for both strings, or their lengths are different, or both. If they have different characters at one or more index positions, let k be the smallest such index; then the string whose character at position k has the smaller value, as determined by using the < operator, lexicographically precedes the other string. In this case, compareTo returns the difference of the two character values at position k in the two string -- that is, the value:

this.charAt(k)-anotherString.charAt(k)

If there is no index position at which they differ, then the shorter string lexicographically precedes the longer string. In this case, compareTo returns the difference of the lengths of the strings -- that is, the value:

this.length()-anotherString.length()