Reputation: 131
I know that compareTo
returns a negative or positive result on how well one string correlates to the other, but then why:
public class Test {
public static void main(String[] args) {
String y = "ab2";
if(y.compareTo("ac3") == -1) {
System.out.println("Test");
}
}
}
is true and
public class Test {
public static void main(String[] args) {
String y = "ab2";
if(y.compareTo("ab3") == -1) {
System.out.println("Test");
}
}
}
is also true?
Upvotes: 13
Views: 30444
Reputation: 137064
The general contract of Comparable.compareTo(o)
is to return
In your example "ab2".compareTo("ac3") == -1
and "ab2".compareTo("ab3") == -1
only means that "ab2"
is lower than both "ac3"
and "ab3"
. You cannot conclude anything regarding "ac3"
and "ab3"
with only these examples.
This result is expected since b
comes before c
in the alphabet (so "ab2" < "ac3"
) and 2
comes before 3
(so "ab2" < "ab3"
): Java sorts Strings lexicographically.
Upvotes: 20
Reputation: 2247
compareTo() compares two string with regard to their alphabetical order. Both of you tests have a String that is alphabetically sorted "before" the String you compare it with. In other words:
By the way, you should rather use "< 0" than "== -1" in your if statement, as the compareTo
spec says that the function returns a negative number, not specifically "-1"
Upvotes: 1
Reputation: 393781
compareTo
for String
s returns -1 if the first String
(the one for which the method is called) comes before the second String
(the method's argument) in lexicographical order. "ab2" comes before "ab3" (since the first two characters are equal and 2 comes before 3) and also before "ac3" (since the first character is equal and b comes before c), so both comparisons return -1.
Upvotes: 1