Hi there gents

 

I managed to find this function which converts a 2 byte hex to 2 byte BCD, but since 2 bytes of hex can hold a decimal value 0 to 255, it is limited to a hex value between 0x00 and 0x63 (0 to 99).

 

uint8_t hex2bcd (uint8_t x)

{

uint8_t y;

y = (x / 10) << 4;

y = y | (x % 10);

return (y);

}

I need some help adapting this to produce a 12 bit (can use a 16bit variable) BCD output.

I have developed my own code, but it is very very long and highly inefficient, so I am looking to use the simplest code and/or fasted code possible.

CK

 

edit: came up with this, which doesn't seem to work.

 

uint16_t hex2bcd12bit (uint8_t x)

{

uint16_t B1;

uint16_t B2;

uint16_t B3;

B1 = x & 0x0F;

B2 = (x & 0xF0) >> 4;

B1 = B1 % 10;

B2 = B2 + B1 / 10;

B3 = B2/10;

B2 = B2 % 10;

B3 = B3<<8 + B2<<4 + B1;

return (B3);

}

Edited April 11, 201510 yr by CasualKilla

y = (x / 100) << 8; // will get 1xx-2xx

but x contain upper 1/2, and you have to get rid of.

x = x % 100; // modulo

Then the rest of original hex2bcd().

Result type will have to be changed.

Edited April 11, 201510 yr by Sensei