Java int to unsigned byte

Question

Long story short, I'm reading some integer values from one file, then I need to store them in a byte array, for later writing to another file.

For example:

int number = 204;
Byte test = new Byte(Integer.toString(number));

This code throws:

java.lang.NumberFormatException: Value out of range. Value:"204" Radix:10

The problem here is that a byte can only store from -127 - 128, so obviously that number is far too high. What I need to do is have the number signed, which is the value -52, which will fit into the byte. However, I'm unsure how to accomplish this.

Can anyone advise?

Thanks

You need to take an array of bytes(usually buffer) .. and have to store each byte — Pragnani Kinnera
– Pragnani Kinnera, Commented May 17, 2013 at 15:10
if you want to preserve integer's value then you cant cast it to byte ? you need to read all the bytes from file. — Peeyush
– Peeyush, Commented May 17, 2013 at 15:17
possible duplicate of How to Convert Int to Unsigned Byte and Back — fglez
– fglez, Commented May 20, 2013 at 13:38

Bernhard Barker · Accepted Answer · 2013-05-17 15:12:02Z

15

A much simpler approach is to cast it:

int number = 204;
byte b = (byte)number;
System.out.println(b);

Prints -52.

answered May 17, 2013 at 15:12

Bernhard Barker

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laz · Accepted Answer · 2013-05-17 15:12:04Z

4

You can cast it:

Byte test = (byte) number;

answered May 17, 2013 at 15:12

laz

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Andy Thomas · Accepted Answer · 2013-05-17 15:14:12Z

1

Use this:

  byte b = (byte) ((0xFF) & number);

answered May 17, 2013 at 15:14

Andy Thomas

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2 Comments

Bernhard Barker Over a year ago

I don't think you need to & it with 0xFF. It seems to output the same with or without it. And yes, I did try with numbers > 255.

Andy Thomas Over a year ago

For the compiler, it's not necessary in Java, for this narrowing conversion from int to byte. However, it explicitly documents that you're intentionally using only the least-significant byte.

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