CODEWITHSUNDEEP
c
jpen
jpen

Reputation: 2147

C - Converting decimal number to packed BCD

How can I achieve these conversions in C without using sprintf?

20 => 0x20
12 => 0x12

Currently I have:

int year = 12;
int month = 10;
int day = 9;
unsigned char date[3];

date[0] = year & 0xFF;
date[1] = month & 0xFF;
date[2] = day & 0xFF;

date will contain { 0x0C, 0x0A, 0x09 } but I want it to be { 0x12, 0x10, 0x09 }

Upvotes: 0

Views: 15306

Answers (3)

aaa
aaa

Reputation: 715

I found the same difficulty when using a RTCC on my PIC mcu. It somehow is common to store a value from 0to99 in a byte and using the lower and higher nybble as a decimal value.

So the binary nybbles from a char might be:

0001 0010 (Binary BCD coded value)
 1    2   (Decimal BCD representation) ^ That would be 12 BCD but 18 Binary

While 00010010 would be 18 when coded in regular binary (https://www.google.nl/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=0b00010010+to+decimal)

I used the code below to fix this.

#define HI_NIBBLE(b) (((b) >> 4) & 0x0F)
#define LO_NIBBLE(b) ((b) & 0x0F)

char BcdToDecimal(char bcd){
    return (char)((HI_NIBBLE(bcd)*10)+(LO_NIBBLE(bcd)));
}

char DecimalToBcd(char decimal){
    return (char) ((decimal / 10)*16)+(decimal % 10);
}

Upvotes: 0

Steve Jessop
Steve Jessop

Reputation: 279255

For the limited 2-digit range you're using:

assert(year >= 0 && year < 100);
date[0] = (year / 10) * 16 + (year % 10);

etc.

You could express it as ((year / 10) << 4) | (year % 10) if that makes more sense to you.

Upvotes: 5

Hola Soy Edu Feliz Navidad
Hola Soy Edu Feliz Navidad

Reputation: 7072

You just have to retrieve each digit in decimal base en multiply it to its equivalent in hexadecinal.

#include <stdio.h>

int hex(int v){
int total = 0;
int resultbase = 1;
while(v > 0 ){
    total += resultbase * (v % 10);
    resultbase *= 16;
    v /= 10;
}

return total;
}

 int  main(){
printf ("12 => %x, 20 => %x\n", hex(12), hex(20));
return 0;

}

Upvotes: 6

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