QUESTION 7:
Decide on the value returned by compareTo():
| "bugbear" | .compareTo | ("bugbear") | zero |
| "bugbear" | .compareTo | ("Bugbear") | NOT zero |
| "Zorba" | .compareTo | ("Zorba!") | NOT zero |
| "mushroom" | .compareTo | ("mush room") | NOT zero |
| "TOAD" | .compareTo | ("TOAD") | zero |
If the strings are not identical, you have to decide which one is less than the other. Sometimes this depends only on the length of the strings.
Rule 2: Otherwise, if string
Ais a prefix of stringB, thenA.compareTo(B) < 0.
"bat".compareTo("batcave") < 0
"apple".compareTo("applesauce") < 0
"BAT".compareTo("BATCAVE") < 0
If it is the first string that is longer, then that string is greater than the other:
"batcave".compareTo("bat") > 0
"applesauce".compareTo("apples") > 0
"BATCAVE".compareTo("BAT") > 0
In using this rule you need to pay attention to case. "BAT" is not a prefix of "batcave".
There are yet more rules ....
Decide on the value returned by compareTo():