How to convert decimal back BCD?

Question

I am able to convert BCD to decimal, for example I can convert 0x11 to 11 instead of 17 in decimal. This is the code I used.

unsigned char hex = 0x11;
unsigned char backtohex ;

int dec = ((hex & 0xF0) >> 4) * 10 + (hex & 0x0F);

Now I want to convert dec back to BCD representation. I want 11 to be converted back to 0x11 not 0x0B. I am kind of confused as how to go back.

Thanks!

You can't use bit manipulation, but division and modulo operators in a loop. — Some programmer dude, Jan 25 '15 at 17:30
Your hex to dec function is flawed; e.g., 0x0A and 0x10 both convert to 10. So there is no way to "go back." Maybe you don't mean hexadecimal at all, just BCD. — Doug Currie, Jan 25 '15 at 17:55

score 4 · Accepted Answer · answered Jan 25 '15 at 17:31

4

Assuming your input is always between 0 and 99, inclusive:

unsigned char hex = ((dec / 10) << 4) | (dec % 10);

Simply take the upper digit and shift it left by one nibble, and or the lower digit in place.

answered Jan 25 '15 at 17:31

hobbs

1 Answers1