equals()
should be the method of choice in the case of the OP.
Looking at the implementation of equals()
and compareTo()
in java.lang.String on grepcode, we can easily see that equals is better if we are just concerned with the equality of two Strings:
equals()
:
1012 public boolean equals(Object anObject) {
1013 if (this == anObject) {
1014 return true;
1015 }
1016 if (anObject instanceof String) {
1017 String anotherString = (String)anObject;
1018 int n = count;
1019 if (n == anotherString.count) {
1020 char v1[] = value;
1021 char v2[] = anotherString.value;
1022 int i = offset;
1023 int j = anotherString.offset;
1024 while (n-- != 0) {
1025 if (v1[i++] != v2[j++])
1026 return false;
1027 }
1028 return true;
1029 }
1030 }
1031 return false;
1032 }
and compareTo()
:
1174 public int compareTo(String anotherString) {
1175 int len1 = count;
1176 int len2 = anotherString.count;
1177 int n = Math.min(len1, len2);
1178 char v1[] = value;
1179 char v2[] = anotherString.value;
1180 int i = offset;
1181 int j = anotherString.offset;
1183 if (i == j) {
1184 int k = i;
1185 int lim = n + i;
1186 while (k < lim) {
1187 char c1 = v1[k];
1188 char c2 = v2[k];
1189 if (c1 != c2) {
1190 return c1 - c2;
1191 }
1192 k++;
1193 }
1194 } else {
1195 while (n-- != 0) {
1196 char c1 = v1[i++];
1197 char c2 = v2[j++];
1198 if (c1 != c2) {
1199 return c1 - c2;
1200 }
1201 }
1202 }
1203 return len1 - len2;
1204 }
When one of the strings is a prefix of another, the performance of compareTo()
is worse as it still needs to determine the lexicographical ordering while equals()
won't worry any more and return false immediately.
In my opinion, we should use these two as they were intended:
equals()
to check for equality, andcompareTo()
to find the lexical ordering.