question: public int compareTo ( java . lang . String anotherString ) { if ( anotherString == this ) { return 0 ; } int minL = Math . min ( anotherString . length ( ) , length ( ) ) ; for ( int iter = 0 ; iter < minL ; iter ++ ) { char a = value [ offset + iter ] ; char b = anotherString . value [ anotherString . offset + iter ] ; if ( a != b ) { return a - b ; } } return length ( ) - anotherString . length ( ) ; }
target:  Compares two strings lexicographically. The comparison is based on the Unicode value of each character in the strings. The character sequence represented by this String object is compared lexicographically to the character sequence represented by the argument string. The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal; compareTo returns 0 exactly when the method would return true. 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()