[SOLVED] Bitwise & operator: what am I missing?

atlantis43 · 03-08-2015, 08:01 PM

In a text description of Bitwise &, it is mentioned:

Quote:

The bitwise AND operator is often used with a mask.-----
One common C usage is this statement:
ch &= 0xff;
This mask leaves the final 8 bits of ch alone and sets the rest to 0. Regardless whether the original ch is 8 bits, 16 bits, or more, the final value is trimmed to something that fits into a single byte.

Though I'm not aware of the exact purpose of such "truncation" in coding, when I run the following:

Code:

#include <stdio.h>
 int main(void)
 {
 	char MASK = 0xff;
  	unsigned int a = 521027;
 	unsigned int b;
	b = a & MASK;
	printf("b is now u\n",b);
 	return 0;
 }

I get the following output, which does not appear "truncated" to an eight bit value:

Quote:

val is now 521027

What am I missing (or misunderstanding)?
Thanks.

Keith Hedger · 03-08-2015, 08:14 PM

use unsigned int for MASK at the moment MASK is a char which is 8 bits but you are appling it to an int which can vary in size depending on the architecture of you machine, and could be anything up to 32 bits long, using tthe same size will mean you wont have undefined bits at the top end.

metaschima · 03-08-2015, 08:21 PM

You want the mask to be 'unsigned char'.

signed char

Code:

bash-4.2$ ./tt
MASK 4294967295, a 521027
b is now 521027

unsigned char

Code:

bash-4.2$ ./tt
MASK 255, a 521027
b is now 67

It has to do with the way the signed integers are represented.

atlantis43 · 03-08-2015, 09:43 PM

Thanks to both replies, I now see it working properly. It seems that the text was misleading in the statement

Quote:

Regardless whether the original ch is 8 bits, 16 bits, or more, the final value is trimmed to something that fits into a single byte.

That was why I tried the code with

Code:

char MASK = 0xff;

expecting that I would see the truncation suggested.
Thanks for your forum, as I always find that so many texts serve better as reference books for those who are already familiar with the subject.

NevemTeve · 03-09-2015, 12:44 AM

You shouldn't mix binary operations and signed numbers. Unsigned is easier, and doesn't have platform specific behaviours.
Now masking the LS byte is goes like this:

Code:

	b = (unsigned char)a;

atlantis43 · 03-09-2015, 07:00 AM

Quote:

Originally Posted by NevemTeve

You shouldn't mix binary operations and signed numbers. Unsigned is easier, and doesn't have platform specific behaviours.
Now masking the LS byte is goes like this:

Code:

	b = (unsigned char)a;

Everything well explained, but what does "LS byte" stand for?

NevemTeve · 03-09-2015, 07:14 AM

(least significant, as in Least significant bit)

atlantis43 · 03-09-2015, 07:34 AM

Quote:

Originally Posted by NevemTeve

(least significant, as in Least significant bit)

Oh! In my readings, the sign-determinant bit of a signed int is shown (at least pictorially) as the N-1 bit. Most confusing.

NevemTeve · 03-09-2015, 07:44 AM

(Ordering bits of a byte (byte of a word) isn't entirely unambigous either. For bits, the cleanest notations might be this: 2^0,2^1,...2^7)