How to convert char to hex stored in uint8_t form?

Question

Suppose I have these variables,

const uint8_t ndef_default_msg[33] = {
    0xd1, 0x02, 0x1c, 0x53, 0x70, 0x91, 0x01, 0x09,
    0x54, 0x02, 0x65, 0x6e, 0x4c, 0x69         


        

Answer 1

char *ndef_input="Z";

uint8_t b=90; //assume this is your character Z in decimal ascii code 90 and HEX = 5A
uint8_t LSB = b & 0x0F; // this is LSB 10 decimal = A
uint8_t MSB = (b & 0xF0)>>4; // this is MSB 5 in decimal = 5 in Hex
cout << "MSB" << MSB << "LSB" << LSB;