Minimum string as per String#CompareTo

旧时模样 提交于 2019-12-01 21:04:59

Given the implementation of String#compareTo (see source) you can verify that a string of size 0 (thus the empty string) will always be inferior or equal to any other non null string. (equal in case of comparing to the empty string).

Edit:

Try this,

   String s="";
      String s1=new String(new char[-2]);  // Here you will get NagativeArraySize Exception
     System.out.println(s1.compareTo(s));

Here compareTo returns nothing in System.out.println(), so the minimum string will be "" or String.Empty


AFAIK, There is no minimum lenght and you can not find a string with the length <0, so the minimum length for the string 0 (string.Empty or "").

Check this source code of CompareTo ,

public int compareTo(String anotherString) {
        int len1 = count;       // original string count
        int len2 = anotherString.count;  // comparision string count
        int n = Math.min(len1, len2);       
        char v1[] = value;
        char v2[] = anotherString.value;
        int i = offset;
        int j = anotherString.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;
    }

There is no minimum length, since you can validly compare two empty strings with "".compareTo("")

Yes it is a empty String with not values in it.

Take a look at this sample code :

String str = "";
str.compareTo("");

Yes the minimum length string will be an empty string "" with a length of 0. I don't think there will be such a string with length less than 0.

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