Comparing unsigned char bits.


 
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# 1  
Old 04-15-2012
Comparing unsigned char bits.

Code:
/******************************************************************************/
/* Printing an unsigned character in bits                                       */

#include    <stdio.h>

void display_bits ( unsigned char );

int main()
{
    unsigned char x;                          /* Number to be printed in bits  */

    printf ( "Enter an unsigned character: " );
    scanf ( "%c", &x );
    display_bits ( x );
}

void display_bits ( unsigned char value )
{
    unsigned char i,
                 display_mask = 1 << ( 8 * sizeof(unsigned char) - 1 );
printf("%c\n",display_mask);
    /* The display_mask contains 1 shifted by the number of bits in int       */

    printf ( "%7c = ", value );

    for ( i = 1; i <= ( 8 * sizeof(unsigned char)); i++ )
    {
        putchar ( ( value & display_mask ) ? '1' : '0' );
        value <<= 1;

        if ( !( i % 8 ) )
            putchar ( ' ' );
    }

    putchar ( '\n' );
}

The output here is not what I am expecting. I want the input to be displayed in binary.
# 2  
Old 04-15-2012
What exactly is not what you are expecting ? Your program output is displayed in binary and the values looks correct to me.

There are minor things that can be improved, like:
  • computing sizeof(unsigned char) which is pointless as it is always 1 by the standard.
  • printf("%c\n",display_mask); doesn't print anything useful as ASCII character 128 is non printable.
  • printf ( "%7c = ", value ); %7c should be just %c as you aren't using wide (multibyte) characters.
# 3  
Old 04-15-2012
If I enter 9, the output is

Enter an unsigned character: 9
?
9 = 00111001

So, the question mark is because 128 is too big for 1 byte and I am not expecting 00111001 because that is 2^5 + 2^4 + 2^3 + 1.
# 4  
Old 04-15-2012
Quote:
Originally Posted by robin_simple
So, the question mark is because 128 is too big for 1 byte
No, bytes are from 0 to 255. The question mark is there because ASCII character 128 is non printable.
Quote:
and I am not expecting 00111001 because that is 2^5 + 2^4 + 2^3 + 1.
Well, 00111001 is precisely the binary representation of the '9' character. What are you expecting instead ?
# 5  
Old 04-15-2012
00001001

---------- Post updated at 03:34 PM ---------- Previous update was at 03:30 PM ----------

How could I translate each binary position in an unsigned char to a corresponding number 1 - 8?
# 6  
Old 04-15-2012
Numbers already are bits, so you can just use the bit shift operator.

Code:
# Get bit
int bit=3; // bit from 0 to 31
unsigned int bitval=(1<<bit);

printf("Value of bit %u is %u\n", bit, bitval);

And do binary math to see when they're set.

Code:
unsigned char value=255;
int bit=3;
unsigned char bitset=value & (1<<bit);

if(bitset) { printf("bit %d set\n", bit+1); }
else       { printf("bit %d unset\n", bit+1); }

# 7  
Old 04-15-2012
Quote:
Originally Posted by robin_simple
00001001
Then read the "9" as a number instead of as a character...
Code:
scanf("%u", &x);

Or read it as a char and extract out the the number by subtracting '0' from it...
Code:
scanf("%c", &x);
x = x - '0';

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