Print 2 numbers stored in 24-bits in decimal format

Question

An implementation function from the nice little Wiegand library YetAnotherArduinoWiegandLibrary prints a hexadecimal representation of a facility # & card # (8-bit facility #, 16-bit card #) — 24 bits of a 26-bit stream that comes across the wire.

Example encoded 24-bit number: 7111097 (decimal). Hex equivalent: 6C81B9. The 6C portion is the facility # (decimal: 108). 81B9 is card # (decimal 33209).

This function from the library's example usage serial prints the hexadecimal format of the number:

void receivedData(uint8_t* data, uint8_t bits, const char* message) {
 Serial.print(message);
 Serial.print(bits);
 Serial.print("bits / ");
 //Print value in HEX
 uint8_t bytes = (bits+7)/8;
 for (int i=0; i<bytes; i++) {
 Serial.print(data[i] >> 4, 16); //1st, 3rd, 5th hex digit
 Serial.print(data[i] & 0xF, 16); //2nd, 4th, 6th hex digit
 }
 Serial.println();
}

What I've tried to do unsuccessfully so far is to loop over the bits to also serial print the two decimal formatted numbers (facility:card) in the final output form: 108:33209

• jsotola: why do you need to loop over bits? ... the for loop already manipulates bytes of data
• micahwittman: @jsotola With or without a for loop how do you output the two separate logical numbers in decimal format?
• jsotola: how do you convert 3 bytes to a single number? .... think about this ... what mathematical process do you follow to convert 4 hundreds , 6 tens and 8 units to 468?
• Duncan C: I wonder why that sample code prints each nibble of a byte separately? Doesn't Serial.print(255, 16); output FF? So couldn't the inside of the for loop just read Serial.print(data[i], 16);

Answer 1

Your code loops from the high order byte to the low order byte, printing each byte as 2 hex digits.

If you want instead to print the first byte as a decimal value, and then the remaining bytes as another decimal value, then do this:

Print byte 0 in base 10:

 Serial.print(data[0]); //first byte
 Serial.print(":");

Then loop through the remaining bytes, building a result as a decimal value, and print that:

unsigned long total = 0;
for (int i=1; i<bytes; i++) //Skip the first byte
{
 total = total << 8; //Shift the previous total by 8 bits. 
 //(The same as multiplying by 256, but faster)
 total += data[i]; //Add in the new value
}
Serial.println(total);

The sample code you posted is written to handle a variable number of bytes. The code above does the same thing with decimal results. It's much simpler if you know you will always have 3 bytes or 4 bytes:

For 3 bytes:

 Serial.print(data[0]); //first byte
 Serial.print(":");
 long total = long(data[1])*256 + data[2]);
 Serial.println(long(total); 

• Duncan C: I wrote the second version of the code using long ints, since I started out with printing an 8-bit value followed by a 24 bit value. If the second value is only 2 bytes (16 bits) you could change the data type to unsigned int
• micahwittman: This is great and makes sense of it for me, thank you.