C how to Shift bits in BCD

**evariste** · 12-16-2010

Hi, i have a binary data that is coded in BCD bytes , example: AB(one BCD byte) in BCD is the concat of A and B in binary, so : AB = 1010(A)1011(B), and i want a function that can reverse the my BCD byte, the result will be :

Code:

shift_function(AB) = 1011(B)1010(A)

Thanks.

**itsme86** · 12-16-2010

So you want the high and low nibbles of the byte?

Code:

int main(void)
{
...
int a, b;
shift_function(bcd_byte, &a, &b);
...
}

void shift_function(int bcd_byte, int *high, int *low)
{
  *high = bcd_byte >> 4;
  *low = bcd_byte & 0xF;
}

**evariste** · 12-16-2010

Thanks for the reply,
i try your code (you shift 4 bits) and get a strange result:

Code:

#include <stdio.h>

void shift_function(unsigned bcd_byte, int *high, int *low)
{
  *high = bcd_byte >> 4;
  *low = bcd_byte & 0xF;
}

int
main()
{
        int a, b;
        unsigned char bcd_byte = 0x34;

        printf("Before : %x\n", bcd_byte);
        shift_function(bcd_byte, &a, &b);
        printf("After : %x\n", bcd_byte);

        return 0;
}

**quzah** · 12-16-2010

What exactly are you expecting to happen? You pass a value bcd_byte to a function, not its address, and you are expecting the original variable to be changed? Why aren't you looking at a's and b's values? The function doesn't actually change bcd_byte. How could it? It's a value pass.

Quzah.

**anduril462** · 12-16-2010

For one, you don't really have BCD. BCD is binary coded decimal, which means each nibble can only store valid decimal values (0-9), so you can't have A and B representing 10 and 11. Regardless, you want to switch the high and low nibbles, so try something like this:

Code:

void swap_nibbles(unsigned char *c)
{
    *c = (*c << 4) | (*c >> 4)
}

int main(void)
{
    unsigned char c = 0xab;
    swap_nibbles(&c);
    return 0;
}

EDIT: Fixed a typo in my code sample. Thanks ncward!

**evariste** · 12-16-2010

Originally Posted by quzah

What exactly are you expecting to happen? You pass a value bcd_byte to a function, not its address, and you are expecting the original variable to be changed? Why aren't you looking at a's and b's values? The function doesn't actually change bcd_byte. How could it? It's a value pass.

Quzah.

You're absolutly right i'm sorry. But this function does not do what i'm expecting.

**evariste** · 12-16-2010

Code:

#include <stdio.h>

unsigned char shift_function(char bcd_byte)
{
  int high, low;

  high = bcd_byte >> 4;
  low = bcd_byte & 0xF;

  return low|high;
}

int
main()
{

        char bcd_byte = 0x34;
        unsigned char result;

        printf("Before : %x\n", bcd_byte);
        result = shift_function(bcd_byte);
        printf("After : %x\n", result);
       
        return 0;
}

I try this code but?

**anduril462** · 12-16-2010

But you forgot to shift low over 4 bits before ORing it with high:

Code:

  low = (bcd_byte & 0xF) << 4;

or just

Code:

  low = bcd_byte << 4;

**quzah** · 12-16-2010

But what? Stop giving cryptic half-sentence responses, and tell us what you expect to see in your output that you aren't seeing.

Quzah.

**evariste** · 12-16-2010

Originally Posted by anduril462

But you forgot to shift low over 4 bits before ORing it with high:

Code:

  low = (bcd_byte & 0xF) << 4;

or just

Code:

  low = bcd_byte << 4;

Thanks anduril462 problem solved.

**evariste** · 12-16-2010

Originally Posted by quzah

But what? Stop giving cryptic half-sentence responses, and tell us what you expect to see in your output that you aren't seeing.

Quzah.

Sorry problem solved thanks.

**ncward** · 12-16-2010

Originally Posted by anduril462

For one, you don't really have BCD. BCD is binary coded decimal, which means each nibble can only store valid decimal values (0-9), so you can't have A and B representing 10 and 11. Regardless, you want to switch the high and low nibbles, so try something like this:

Code:

void swap_nibbles(unsigned char *c)
{
    *c = (*c << 4) && (*c >> 4)
}

int main(void)
{
    unsigned char c = 0xab;
    swap_nibbles(&c);
    return 0;
}

Code:

void swap_nibbles(const char *c)
     *c = (*c << 4) | (*c >> 4);

I think is what you meant.

**anduril462** · 12-16-2010

D'oh! Yeah, that's it. Thanks.

**evariste** · 12-17-2010

Originally Posted by ncward

Code:

void swap_nibbles(const char *c)
     *c = (*c << 4) | (*c >> 4);

I think is what you meant.

Thanks, TOPIC is closed

.

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