- C
The bitwise OR in C
Hi there, I'm new in C and using a downloaded tutorial to practice and learn. Heading up to the 17th chapter I have a question I can't find an answer to.
The exercise is showing me the use of the bitwise OR:
What do i have to change to let the bitwise OR give a result higher then 127?
The exercise is showing me the use of the bitwise OR:
#include <stdio.h>
#define SET 32
char *binbin(int n);
int main()
{
int bor, result;
printf("Type a value from 0 to 255: ");
scanf("%d",&bor);
result = bor | SET;
printf("\t%s\t%d\n",binbin(bor),bor);
printf("|\t%s\t%d\n",binbin(SET),SET);
printf("=\t%s\t%d\n",binbin(result),result);
return 0;
}
char *binbin(int n)
{
static char bin[9];
int x;
for (x=0;x<8;x++)
{
bin[x] = n & 0X80 ? '1' : '0';
n << 1;
}
bin[x]='\0';
return(bin);
}
Type a value from 0 to 255: 95
01011111 95
| 00100000 32
= 01111111 127
Type a value from 0 to 255: 96
01100000 96
| 00100000 32
= 01100000 96
printf("This is the result from bitwise OR: %d\n",result);
result = bor | SET;
What do i have to change to let the bitwise OR give a result higher then 127?
Hi,
I'm not sure if this is the error, on my compiler (I tested in C++) the exact same code produces some other output - but there's one line which makes no sense.
Hope that helps,
ZOPPO
I'm not sure if this is the error, on my compiler (I tested in C++) the exact same code produces some other output - but there's one line which makes no sense.
n << 1;
n = n << 1;
// or
n <<= 1;
static char bin[9];
Hope that helps,
ZOPPO
0 OR 1 = 1
1 OR 1 = 1
1 OR 0 = 1
96 = 64 + 32 so any value from 96 to 127 will have 2^5 bit set 00010000
128 + 32 = 160 - > 255 so 160 to 255 will also have 2^5 set
1 OR 1 = 1
1 OR 0 = 1
96 = 64 + 32 so any value from 96 to 127 will have 2^5 bit set 00010000
128 + 32 = 160 - > 255 so 160 to 255 will also have 2^5 set
@ozo : I didn't, but reading the comment of David, I understand that a value between 128 and 160 also does the trick.
@ZOPPO : That was a typo indeed. I changed it before I displayed the outcome in my question. Thanks for that!
@David : Thanks, but how do I change the code so the value from 96 to 127 will give the correct outcome? Or is that just not possible?
@ZOPPO : That was a typo indeed. I changed it before I displayed the outcome in my question. Thanks for that!
@David : Thanks, but how do I change the code so the value from 96 to 127 will give the correct outcome? Or is that just not possible?
Hm - I did read it again, and I'm not sure what's it what you think is wrong.
95 | 32 = 0x01011111 | 0x00100000 = 0x01111111 = 127
96 | 32 = 0x01100000 | 0x00100000 = 0x01100000 = 96
97 | 32 = 0x01100001 | 0x00100000 = 0x01100001 = 97
...
@Zoppo: I think I did expect the reaction I got from entering 95. So entering 96 would have resulted in 128...I do understand that this is not gonna happen :-)
I'm learning, learning, learning... Thanks!
I'm learning, learning, learning... Thanks!
You're welcome, I'm glad I could help ...
BTW: it's sometimes helpful to have a tool which shows different representations of a number (dec, hex, bin) ... i.e. in Windows the calculator does this in Programmer-Mode.
good luck und best regards,
ZOPPO
BTW: it's sometimes helpful to have a tool which shows different representations of a number (dec, hex, bin) ... i.e. in Windows the calculator does this in Programmer-Mode.
good luck und best regards,
ZOPPO
you seem to think that the OR is adding 32 to any value since the OR'ing does add the 32 bit UNLESS that bit is already set then it does nothing.
when dealing at the bit level you have to think in binary and not in decimal.
000 OR 011 = 011 000 AND 011 = 000
001 OR 011 = 011 001 AND 011 = 001
010 OR 011 = 011 010 AND 011 = 010
011 OR 011 = 011 011 AND 011 = 011
100 OR 011 = 111 100 AND 011 = 000
101 OR 011 = 111 101 AND 011 = 001
110 OR 011 = 111 110 AND 011 = 010
111 OR 011 = 111 111 AND 011 = 011
6 * 2 can be represented as 2 + 2 + 2 + 2 + 2 + 2
6 / 2 can be represented as 6 -2 -2 -2 we keep a count of how many 2's in this case it is 3 before we hit 0 or even quicker just right shift 110 >> 011 = 3
Actually early computers only had a compare if or not 0 and an adder Subtraction was by invert and add 1 and then add the 2 items together
5 - 3 = 101 - 011 (transformed to 100 then add 1 = 101 ) 101 + 101 = 1010 and we ignore the overflow 010 = 2
when dealing at the bit level you have to think in binary and not in decimal.
000 OR 011 = 011 000 AND 011 = 000
001 OR 011 = 011 001 AND 011 = 001
010 OR 011 = 011 010 AND 011 = 010
011 OR 011 = 011 011 AND 011 = 011
100 OR 011 = 111 100 AND 011 = 000
101 OR 011 = 111 101 AND 011 = 001
110 OR 011 = 111 110 AND 011 = 010
111 OR 011 = 111 111 AND 011 = 011
6 * 2 can be represented as 2 + 2 + 2 + 2 + 2 + 2
6 / 2 can be represented as 6 -2 -2 -2 we keep a count of how many 2's in this case it is 3 before we hit 0 or even quicker just right shift 110 >> 011 = 3
Actually early computers only had a compare if or not 0 and an adder Subtraction was by invert and add 1 and then add the 2 items together
5 - 3 = 101 - 011 (transformed to 100 then add 1 = 101 ) 101 + 101 = 1010 and we ignore the overflow 010 = 2
If you want result to be higher then 127, then you must have either bor higher than 127 or SET higher than 127.
Without the static, the
return(bin);
would return an address that was no longer in a valid scope, so the result of
printf("\t%s\t%d\n",binbin
would be undefined.
But I agree that this use of a static address would be poor practice if the function is to be used in a multi-threading application.
Without the static, the
return(bin);
would return an address that was no longer in a valid scope, so the result of
printf("\t%s\t%d\n",binbin
would be undefined.
But I agree that this use of a static address would be poor practice if the function is to be used in a multi-threading application.
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