signed to unsigned. How does this work?

Why/How does this work?

I know that if I want to convert a signed byte (ie 0x7F) to unsigned
number I can promote it to an integer, like this:

byte a = (byte)0xAB; // -85
int b = a & 0xFF; // 171

The bit pattern is 10101011 (0xAB)
If I AND it with 11111111 (0xFF)
The result is 10101011

ie. the same number! so why does this change the number to an unsigned
(positive) number?

re: signed to unsigned. How does this work?

Robert Smith wrote:It is not the same number. You left out all the high bits where the
difference is apparent.

a is a byte, b is an int.

a promotes to int in order to participate in the mask operation; the promoted
value is
0xFFFFFFAB.

Because you masked out the high bits of a's promoted value on purpose, b is
0x000000AB.

Quite different.
Because you masked out all the high bits, including the sign bit, on purpose.

0xFF is an int, and a positive one at that.

0xFF == 0x000000FF

The result of the & is 0x000000AB.

--
Lew

re: signed to unsigned. How does this work?

Thanks so much for explaining that

"Lew" <lew@nospam.lewscanon.comwrote in message
news:LcGdnRlLK8aWzMnbnZ2dnUVZ_vWtnZ2d@comcast.com. ..promotedispurpose.

re: signed to unsigned. How does this work?

Robert Smith wrote:Your avoidance of top-posting in the future will be thanks aplenty.

--
Lew