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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Vidhi Dave
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Amit Shinde
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4Don't store them in a byte. Use short or int. – Eran May 30 '18 at 06:44
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Well, I guess you shouldn't use a byte :P – briosheje May 30 '18 at 06:44
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You are storing in int and typecasting it to byte. This is what you will get. – Rohit5k2 May 30 '18 at 06:44
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Store in b/127 no of array of bytes. And sum all at last – Koustuv Ganguly May 30 '18 at 06:45
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Why you are typecasting in byte.. explain purpose briefly – Vidhi Dave May 30 '18 at 06:45
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Please don't downvote on the grounds of obviousness; that is subjective. – Bathsheba May 30 '18 at 06:47
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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 May 30 '18 at 06:51
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1Possible duplicate of [Java code To convert byte to Hexadecimal](https://stackoverflow.com/questions/2817752/java-code-to-convert-byte-to-hexadecimal) – deathangel908 May 30 '18 at 06:53
3 Answers
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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
Andy Turner
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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.
Bathsheba
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