(JAVA)Why we got an integer when we print number in hexadecimal notation?

Question

enter image description hereExample: sout(OxBD); Output: 189

I don't really understand why does compiler converts different representation of a number to integer. Why not boolean? When I try to cast to double(boolean was here, wrote by mistake...) I got an error, also don't know why. With char it works. I attached the photo below.

There is context missing. What are you trying to print? Is "OxBD" a string? Is there a variable you're printing something from? — Leaderboard, Jul 17 '22 at 18:05
Im just learning the language in depth. Just curious about it. — allrightallrightallright, Jul 17 '22 at 18:06
You can indeed edit your post; while you can't embed, the website will automatically link to imgur. — Leaderboard, Jul 17 '22 at 18:10
@allrightallrightallright `0xBD` is the hexadecimal integer literal for the number `189`. The "value" of `0xBD` is still an integer. — Progman, Jul 17 '22 at 18:13
Sorry, I made a mistake; I copied "OxBD" instead of "0xBD" (as that is what your post said as of writing). You are right @allrightallrightallright. I have deleted my earlier comment. — Leaderboard, Jul 17 '22 at 18:13
Does this answer your question? [Why are hexadecimal numbers prefixed with 0x?](https://stackoverflow.com/questions/2670639/why-are-hexadecimal-numbers-prefixed-with-0x) — Progman, Jul 17 '22 at 18:15
Do you mean why doesn't the value print as hex? How it prints has nothing to do with how you assign a value. — Nathan Hughes, Jul 17 '22 at 18:16
Not really. I don't understand why compiler converts the 0xBD to integer and gives me the result as int... — allrightallrightallright, Jul 17 '22 at 18:16
Oh wow, after reading "hexadecimal integer" it gave more sense to me. — allrightallrightallright, Jul 17 '22 at 18:20
More about it: [JLS § 3.10.1 – integer literals](https://docs.oracle.com/javase/specs/jls/se18/html/jls-3.html#jls-3.10.1). — MC Emperor, Jul 17 '22 at 20:20

score 0 · Accepted Answer · answered Jul 17 '22 at 18:15

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"0xBD" is an integer, which explains why it's converted to hexadecimal by Java. You can't convert to boolean, as 189 isn't something that can be converted to boolean (which can store only two values, 0 and 1).

189 can be stored as a char though; what you get is the ASCII character representation of 189.

answered Jul 17 '22 at 18:15

Leaderboard

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0xBD is an integer? So basically it's just a representation of an integer? – allrightallrightallright Jul 17 '22 at 18:19
Indeed, just in a different base (base 16 instead of base 10). This is also called as hexadecimal. – Leaderboard Jul 17 '22 at 18:20
But still a bit confused why compiler converts to decimal(?). Why it cannot just print in hexadecimal? – allrightallrightallright Jul 17 '22 at 18:21
Because Java recognises that 0xBD is an integer, and (essentially) converts it to its decimal equivalent before printing. It doesn't have a "since I initialised as a hexadecimal, print it only in hexadecimal" mode. – Leaderboard Jul 17 '22 at 18:23
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`System.out.println`, when applied to an `int`, always prints it in decimal. – Louis Wasserman Jul 17 '22 at 18:36

(JAVA)Why we got an integer when we print number in hexadecimal notation?

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