Java converting byte to char - conversion questions

Question

I found this solution for converting a byte to a char on StackOverflow:

byte b = 65;
char c = (char)(b & 0xFF);

What i don't get is the bitwise and with 0xFF. How is the above different from this one?

byte b = 65;
char c = (char)b;

Can someone explain to me step by step what gets converted to what type?

Duplication http://stackoverflow.com/questions/17912640/byte-and-char-conversion-in-java — Sergei Rybalkin, Sep 21 '16 at 22:06

score 1 · Accepted Answer · answered Sep 21 '16 at 22:13

"& 0xff" is typically used in Java to cut off the sign extension for implicit conversions from bytes to larger integer types.

In Java, byte is a signed data type, where values are stored using two's complement, so -1 is represented as 0xff. The corresponding int representation of -1 is 0xffffffff. Thus, when converting from a byte to an integer, all additional leading bits are filled from the highest bit of the byte.

The bitwise and is not needed for byte values up to 0x7f. But if you want to convert a e.g. a non-breaking space (code point 160 / 0xa0), it would get sign extended to 0xffa0, which is a different character than the intended char value of 0x00a0.

I think the thing that threw me off was that 0xFF is an int and not a byte. — Luca Fülbier, Sep 21 '16 at 22:30

Java converting byte to char - conversion questions

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