关于compareTo方法到底是怎么比较顺序的？

[a227192005]

import java.util.*;

public class Test {
public static void main(String args[]) {
List t = new LinkedList();
t.add(new Name("Karl", "M"));
t.add(new Name("John", "O"));
t.add(new Name("Tom", "M"));
System.out.println(t);
Collections.sort(t);
System.out.println(t);
}
}

class Name implements Comparable{
String firstName, lastName;
Name(String firstName, String lastName) {
this.firstName = firstName;
this.lastName = lastName;
}

public String toString() {
return firstName + " " + lastName;
}

public int compareTo(Object o) {
Name n = (Name)o;
int lastCmp = lastName.compareTo(n.lastName);
return
(lastCmp != 0 ? lastCmp : firstName.compareTo(n.firstName));
}
}

[a227192005]

public int compareTo(Object o) {
Name n = (Name)o;
int lastCmp = lastName.compareTo(n.lastName);
return
(lastCmp != 0 ? lastCmp : firstName.compareTo(n.firstName));
}
把引用对象t传进来，强制转换成Name, lastName是当前对象具体指的是哪个，n.lastName又指哪一个

[sunyiz]

public int compareTo(Object o) 这个方法是为了实现Comparable而存在的
而实现Comparable接口是为了可以直接用Arrays.sort()方法对你的这种对象排序

compareTo是将当前对象与参数 o 这个对象比
如果你觉得这个时候应该是当前对象大，就返回个正值
如果你觉得这个时候应该是 o 大，就返回个负值
如果你觉得这个时候应该是一样大，就返回个0

这个方法究竟怎么写，怎么返回，完全是看你觉得这个类的对象应该怎么来比大小

[huntor]

@Override public int compareTo(Object o){
    if (o == this) return true;
    if (o == null) return false;
    if (getClass() != o.getClass()) return false;
    Name that = (Name)o;
    int lc = lastName.compareTo(that.lastName);
    if(lc != 0){
	return lc;
    }
    else{
	return firstName.compareTo(that.firstName);
    }
}

[a227192005]

就拿我这个例子来说， firstName.compareTo(n.firstName)) 总的拿两个对象比较吧，具体是哪两个了？

[sunyiz]

你的这个例子是把当前对象的firstName这个字符串
和参数 o 中的firstName字符串
用String的compareTo方法进行比较的
String具体怎么比大小可以看看String类的源码

[a227192005]

感觉没对上我的问题，，比如“C”，“B”，“A”，三个对象排序firstName.compareTo(n.firstName))，他么具体是怎么比较的？ C.compareTo(C),C.compareTo(B),C.compareTo(A),B.compareTo(C).....这样比的吗？然后得出ABC？

[sunyiz]

public int compareTo(Object o) {
//把Object o转成Name型，便于比较
Name n = (Name)o;
//我们Name怎么比大小呢？那就直接比比我们的lastName吧，
//怎么比，直接用String的compareTo方法好了，java写好的，不用白不用
int lastCmp = lastName.compareTo(n.lastName);
//看看吧如果lastName比出结果了，那最好罗，如果比出来一样咋办？那我们再比比firstName吧
return
(lastCmp != 0 ? lastCmp : firstName.compareTo(n.firstName));  
}  

[sunyiz]

如果楼主想知道java中
Arrays.sort()是怎么通过compareTo比大小的
那看看Arrays中的相关方法吧，下面贴一段里面的代码

    private static void mergeSort(Object[] src,
				  Object[] dest,
				  int low,
				  int high,
				  int off) {
	int length = high - low;

	// Insertion sort on smallest arrays
        if (length < INSERTIONSORT_THRESHOLD) {
            for (int i=low; i<high; i++)
                for (int j=i; j>low &&
			 ((Comparable) dest[j-1]).compareTo(dest[j])>0; j--)
                    swap(dest, j, j-1);
            return;
        }

        // Recursively sort halves of dest into src
        int destLow  = low;
        int destHigh = high;
        low  += off;
        high += off;
        int mid = (low + high) >>> 1;
        mergeSort(dest, src, low, mid, -off);
        mergeSort(dest, src, mid, high, -off);

        // If list is already sorted, just copy from src to dest.  This is an
        // optimization that results in faster sorts for nearly ordered lists.
        if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) {
            System.arraycopy(src, low, dest, destLow, length);
            return;
        }

        // Merge sorted halves (now in src) into dest
        for(int i = destLow, p = low, q = mid; i < destHigh; i++) {
            if (q >= high || p < mid && ((Comparable)src[p]).compareTo(src[q])<=0)
                dest[i] = src[p++];
            else
                dest[i] = src[q++];
        }
    }