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Java Java Arrays Gotchas and Wins Sorting

Posted by Dylan Cascio
Dylan Cascio
5,488 Points

Why does `"Apple".compareTo("apple");` return -32 and not 1, 0 or -1?

As the title says "Apple".compareTo("apple"); returns -32. Similarly, "apple".compareTo("Apple"); returns 32.

Why is this different from the described return values of 1, 0, and -1?

For reference, "APPLE".compareTo("apple"): returns -32 and "apple".compareTo("APPLE"); also returns 32.

1 Answer

Tonnie Fanadez
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Tonnie Fanadez
UX Design Techdegree Graduate 22,796 Points
Tonnie Fanadez
.a{fill-rule:evenodd;}techdegree seal-36
Tonnie Fanadez
UX Design Techdegree Graduate 22,796 Points

Hello Dylan Cascio

Good Question.

The compareTo()method defines the natural order - the default way for ordering objects of a class. The comparison is based on the Unicode Value of each character in the Strings being compared.

Please refer to this link for Unicode Character Table - http://sticksandstones.kstrom.com/appen.html

You will note that the Unicode value for "a" is 097 and "A" is 065 .... "b" is 98 and "B" is 66. and so forth. Since compareTo() is subtracts one value from another, the difference between a Capital Letter and a Lowercase Letter is always -32 for the same letter. The difference is always +32 for (Lowercase - Uppercase) when dealing with the same letter.

It is not a must for 1 or -1 to be returned - unless the letters being tested follow each other in case of "a" and "b" or "c" and "d" or "y" and "z"- Java documentation don't specify it is 1 or -1 to be returned but rather says a positive or a negative integer. 0 is only returned if objects are equal e.g. "d"=="d"

Hope this helps.

Cheers

Dylan Cascio
Dylan Cascio
5,488 Points

Thank you! That's so much clearer now. I see now that I just hadn't written out enough examples in jshell. It looks like even something like "apple".compareTo("orange"); evaluates to -14 and "apple".compareTo("czzzzzzz"); returns -2.