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If I have the following:

byte[] byteArray = new byte[] {87, 79, 87, 46, 46, 46};

I know that the size of each element would be one byte. But what I don't seem to understand is how would the integer 87 be stored in one byte? Or, how does the byte[] store data?

EDIT: I see that you can store -128 to 127 in a byte here in java. So, does that mean there is no way to store anything greater than or lesser than those numbers in a byte[]? If so, doesn't that limit the use of this? Or am not understanding the exact places to use a byte[].

Remy Lebeau
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noMAD
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  • what is different about `87`? even using _signed_ bytes, you can represent up to `127` – pb2q Jul 31 '12 at 19:11
  • I don't get your question. `byte` has a range of -128 to 127 in Java so you can easily store the decimal value 87 in it. – nkr Jul 31 '12 at 19:11
  • Because Java's `byte` type is signed, 8-bit values between 128 and 255 are represented as negative values between -128 and -1, respectively. – Remy Lebeau Jul 31 '12 at 19:38

6 Answers6

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A byte is 8 bits. 2^8 is 256, meaning that 8 bits can store 256 distinct values. In Java, those values are the numbers in the range -128 to 127, so 87 is a valid byte, as it is in that range.

Similarly, try doing something like byte x = 200, and you will see that you get an error, as 200 is not a valid byte.

Jon Newmuis
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  • A `byte` can hold 8-bit values 128-255. It has to be able to, in order to be compatible with other languages that use unsigned bytes instead of signed bytes like Java does. 200 can be represented in a Java `byte` as -56, as both 200 and -56 are represented in 8-bit as `0xC8` hex, `11001000` binary. – Remy Lebeau Jul 31 '12 at 19:35
  • Well, sure, but you can't just directly make such an assignment; you'll have to do some kind of conversion. It would probably be easier to just subtract `128` from your bytes in the range `0...255` to get them into Java's range of `-128...127`. – Jon Newmuis Jul 31 '12 at 19:46
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A byte is just an 8-bit integer value. Which means it can hold any value from -2^7 to 2^7-1, which includes all of the number in {87, 79, 87, 46, 46, 46}.

An integer in java, is just a 4-byte integer, allowing it to hold -2^31 to 2^31 - 1

Rob Wagner
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A Java byte is a primitive with a minimum value of -128 and a maximum value of 127 (inclusive). 87 is within the allowed range. The byte data type can be useful for saving memory in large arrays, where the memory savings actually matters.

A byte[] is an Object which stores a number of these primitives.

Reimeus
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I think the short answer is that byte[] stores bytes. The number 87 in your array above it a byte, not an int. If you were to change it to 700 (or anything higher than 127) you'd get a compile error. Try it.

Mason Bryant
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You can use byte to store values of 8 bit in it which have a (signed) range from from -128 to 127.

With byte[] you can do some special operations like building Strings from a given bytestream and decode them with a desired Charset, and some functions will give you byte[] as their return value.

I don't know enough about the internals of the JVM but it might save memory though.

nkr
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this is because, the computer stores values in a circular progression. not a linear progression like we learn in mathematics. it is because the memory is not infinite. it is finite. so every data type is storing values as a circular progression. to learn more go to this link and read the article.

https://medium.com/@hmsathyajith/numbering-system-edge-cases-of-java-237377553444

Sathyajith Saliya
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