java - Java中compareTo的逻辑(offset的作用)

Question

public class CustomString {
    public char value[];
    public int offset;
    public int count;
    public int hash;
}


CustomString one = new CustomString();
char valueOne[] = {'A','N'};
one.count = 2;
one.hash = 0;
one.offset = 0;
one.value = valueOne;

CustomString two = new CustomString();
char valueTwo[] = {'F','A','N'};
two.count = 3;
two.hash = 0;
two.offset =1;
two.value = valueTwo;

compareTo(one,two)

字符串的compareTo方法:

public static int compareTo(CustomString one, CustomString two) {
    int len1 = one.count;
    int len2 = two.count;
    int n = Math.min(len1, len2);
    char v1[] = one.value;
    char v2[] = two.value;
    int i = one.offset;
    int j = two.offset;

    if (i == j) {
        int k = i;
        int lim = n + i;
        while (k < lim) {
            char c1 = v1[k];
            char c2 = v2[k];
            if (c1 != c2) {
                return c1 - c2;
            }
            k++;
        }
    } else {
        while (n-- != 0) {
            char c1 = v1[i++];
            char c2 = v2[j++];
            if (c1 != c2) {
                return c1 - c2;
            }
        }
    }
    return len1 - len2;
}

由于对于“FAN”我使用偏移量为1，我认为“FAN”的“AN”将与“AN”进行比较并返回0。但从compareTo开始就没有了。的String返回return len1 - len2;

我的问题是，offset的目的是什么？在compareTo方法？偏移量始终为 0。您能否举个例子，其中一个或两个都具有不同的偏移量？

Answer 1

如果这是为了模仿 Java7 之前的 java.lang.String，则 offset 字段用于实现恒定时间 substring() 操作。请参阅Time complexity of Java's substring()

为了将此与您的示例联系起来，"FAN".substring(1, 3) 会将 count 设置为 2，而不是 3:

CustomString two = new CustomString();
char valueTwo[] = {'F','A','N'};
two.value = valueTwo;
two.offset = 1;
two.count = 2; // <------- the number of characters in the string "AN"

因此，one.compareTo(two) 的计算结果为 0(因为 len1 == len2 == 2)。