Question about bitwise operators

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Question about bitwise operators
10 answers - 449 bytes -
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
#include <stdio.h>
int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;
printf("The value is: %u \n", c);
return 0;
}
The value is: 166
$
What am I missing here?
Thanks in advance
Chad
No.1 | | 225 bytes | |
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
Try it with 256 instead.
678 is 0x2a6
0xa6 is 166
No.2 | | 694 bytes | |
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.

#include <stdio.h>

int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;

1010100110 (678)
0011111111 (255)
AND
0010100110 (166)

printf("The value is: %u \n", c);

return 0;

}
The value is: 166
$

What am I missing here?

That ANDing with 255 is equivalent to taking a number modulo 256, not 255.

678 % 256 = 166, and repeated subtraction of 256 (678 - 256 - 256) will
indeed give you 166.

S.
No.3 | | 373 bytes | |
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.

678 in binary = 1010100110
255 in binary = 0011111111
& together = 0010100110 = 166

& = The bits in the result at set to 1 if the corresponding bits in the
two operands are both 1.
No.4 | | 244 bytes | |

pete wrote:
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
Try it with 256 instead.
678 is 0x2a6
0xa6 is 166
No.5 | | 255 bytes | |
Chad wrote:
pete wrote:
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
Try it with 256 instead.
678 is 0x2a6
0xa6 is 166
No.6 | | 255 bytes | |
Chad wrote:
pete wrote:
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
Try it with 256 instead.
678 is 0x2a6
0xa6 is 166
No.7 | | 1633 bytes | |
Chad wrote:
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.

Why you think 'a & b' should mean 'a - b - b' is a mystery.

#include <stdio.h>

int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;

printf("The value is: %u \n", c);
^^^
%u is the specifier for an unsigned int, not an unsigned long (%lu)

return 0;

}
The value is: 166
$

What am I missing here?

Any understanding of logical operators? See the version below, with a
note following.

#include <stdio.h>

int main(void)
{
unsigned a = 678;
unsigned ff = 255;
unsigned c = a & ff;
char format[] = "value of %6s is %#011o (octal), %010u (decimal)\n";

printf(format, "a", a, a);
printf(format, "ff", ff, ff);
printf(format, "a & ff", a & ff, a & ff);
printf(format, "c", c, c);

return 0;

}

value of a is 00000001246 (octal), 0000000678 (decimal)
value of ff is 00000000377 (octal), 0000000255 (decimal)
value of a & ff is 00000000246 (octal), 0000000166 (decimal)
value of c is 00000000246 (octal), 0000000166 (decimal)

[note]
considering the rightmost 4 octal digits
01 & 00 is binary 001 & 000, obviously 000 (0 octal)
02 & 03 is binary 010 & 011, obviously 010 (2 octal)
04 & 07 is binary 100 & 111, obviously 100 (4 octal)
06 & 07 is binary 110 & 111, obviously 110 (6 octal)

so 01245 & 0377 is 0246 or 166 decimal.
No.8 | | 816 bytes | |
Chad a :
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.

#include <stdio.h>

int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;

printf("The value is: %u \n", c);

return 0;

}
The value is: 166
$

What am I missing here?

bitwise operators are about bits. Hence you should use the octal or the
hexadecimal representation that makes it far more clear.

#include <stdio.h>

int main(void)
{
unsigned long a = 0x2A6;
unsigned long ff = 0xFF;
unsigned long c = a & ff; /* I expect 0xA6 */

printf ("The value is: 0x%X\n", c);

return 0;
}

The value is: 0xA6

Neat.
No.9 | | 817 bytes | |
Chad a :
For some strange reason, I thought the following would give me a value
of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.

#include <stdio.h>

int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;

printf("The value is: %u \n", c);

return 0;

}
The value is: 166
$

What am I missing here?

bitwise operators are about bits. Hence you should use the octal or the
hexadecimal representation that makes it far more clear.

#include <stdio.h>

int main(void)
{
unsigned long a = 0x2A6;
unsigned long ff = 0xFF;
unsigned long c = a & ff; /* I expect 0xA6 */

printf ("The value is: 0x%lX\n", c);

return 0;
}

The value is: 0xA6

Neat.
No.10 | | 809 bytes | |
7 Dec 2005 06:26:33 -0800, "Chad" <cdalten@gmail.comwrote:

>For some strange reason, I thought the following would give me a value
>of 168, not 166. My reasoning was that 678 - 255 - 255 = 168.
>
>#include <stdio.h>
>
>int main(void) {
unsigned long a = 678;
unsigned long ff = 255;
unsigned long c = a & ff;

printf("The value is: %u \n", c);

In addition to the other advice, %u requires an unsigned int. You are
passing an unsigned long. This is undefined behavior.

return 0;
>
>}
>The value is: 166
>$
>
>What am I missing here?
>
>Thanks in advance
>Chad

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