signed to unsigned. How does this work?

[Robert Smith]

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?

[Lew]

Re: signed to unsigned. How does this work?

Robert Smith wrote:

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!

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.

so why does this change the number to an unsigned (positive) number?

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

[Robert Smith]

Re: signed to unsigned. How does this work?

Thanks so much for explaining that

"Lew" <lew@nospam.lew scanon.comwrote in message
news:LcGdnRlLK8 aWzMnbnZ2dnUVZ_ vWtnZ2d@comcast .com...

Robert Smith wrote:

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!

>
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.
>

so why does this change the number to an unsigned (positive) number?

>
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

[Lew]

Re: signed to unsigned. How does this work?

Robert Smith wrote:

Thanks so much for explaining that

Your avoidance of top-posting in the future will be thanks aplenty.

--
Lew