Why is converting two bytes just with shifting is incocrrent?

Question

I read this topic about shifting. I thought that if I have two bytes:

byte hi = //...
byte low = //...

why can't I just do that

short s = (short)((hi << 8) | low)

Why is it incorrect? I thought we 8 bits left shift most significant and leave least significant byte as is. And then just bitwise or them.

you should use ByteBuffer for that like: ByteBuffer.wrap(valBytes).order(ByteOrder.LITTLE_ENDIAN).getShort(); — Fady Saad, Aug 13 '17 at 03:34
@FadySaad I'm aware of `ByteBuffer`. I'm interested in understanding bit-shift operations... just that :) — St.Antario, Aug 13 '17 at 03:35

score 2 · Accepted Answer · answered Aug 13 '17 at 03:52

This gives a wrong result because bytes are signed and extended to int to do the calculations. So take for example

hi = (byte)0x01;
low = (byte)0x80;

then you calculate:

0x00000100 | 0xffffff80 -> 0xffffff80

which is not the desired result.

You could write it like this instead:

short s = (short)((hi << 8) | (low & 0xff))

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