How store values greater than 127 in byte datatype in java.
int b = 160;
System.out.println((byte)b);
It prints -96.
Note : I want to write bytes on a BLE device. So can not convert it to short or int.
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4Don't store them in a byte. Use short or int.Eran– Eran2018-05-30 06:44:01 +00:00Commented May 30, 2018 at 6:44
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Well, I guess you shouldn't use a byte :Pbriosheje– briosheje2018-05-30 06:44:31 +00:00Commented May 30, 2018 at 6:44
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You are storing in int and typecasting it to byte. This is what you will get.Rohit5k2– Rohit5k22018-05-30 06:44:32 +00:00Commented May 30, 2018 at 6:44
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1@Vishva Dave, I want to write bytes on a BLE device. So can not convert it to short or int.Amit Shinde– Amit Shinde2018-05-30 06:51:02 +00:00Commented May 30, 2018 at 6:51
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1Possible duplicate of Java code To convert byte to Hexadecimaldeathangel908– deathangel9082018-05-30 06:53:29 +00:00Commented May 30, 2018 at 6:53
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3 Answers
You might want to store a value in the range 128-255 in a byte. You can, provided you don't also want to store a value -128 to -1 in the same byte (at a different time, obviously).
Just use the bitwise and operator when you want to read it:
b & 0xff
Comments
The range of a byte is the inclusive range -128 to +127.
You can't store a number greater than 127 in a byte, although you can abuse the range -128 to -1. Your output -96 is attained since the eight bit pattern of -96 matches 160 written in binary. Crudely, you can recover your original number if you add 256 to any negative, or mask all apart from the final 8 bits using b & 256.
If you want to store 160 then use a char (0 to 65535), a short, or an int.
Comments
For this reason use int or long.
1 Comment
Eugene
this is more a comment rather than an answer