PIC18F4550 : How to convert 16 bits integer to two 8 bits chars in C?

Question

I'm blocked because of an issue that is causing a lot of time loss. I've created a function to get the LSB and another one to get the MSB.

char get_LSB (int n){
    
    unsigned char lsb = (unsigned)n & 0xFF; // mask the lower 8 bits
   return lsb;
}

char get_MSB (int n){  
    unsigned char msb = ((unsigned)n >> 8 );// shift the higher 8 bits
   return msb;
}

It doesn't work as when I call the function get_LSB or get_MSB I get a result different from what I expected. Let's check the following example:

void main() {
              TRSISC = 0x00;
              PORTC = get_LSB(100);// function should return 100 but returns 68
              Delay_ms(5000);
              PORTC = get_MSB(360); // should return 1 but returns 0
}

Answer 1

Try these:

unsigned char get_LSB (int n)
{    
    unsigned char lsb = (unsigned char) (n & 0xFF); // mask the lower 8 bits
    return lsb;
}

unsigned char get_MSB (int n)
{  
    unsigned char msb = (unsigned char) (n >> 8);// shift the higher 8 bits
    return msb;
}

Answer 2

Using PORTC to check the functionality of your function is going to cause multiple issues.

For example it is a 7 bit register with bit 3 missing. Then RC4 and RC5 are not under complete control of the PORTC alone, the USB peripheral has control over it. And they can only be used as input, so the state depends on the USB or the voltage at the pin, but never on what you write to the PORTC register.

So that means, that bits 3, 4 and 5 are all not what you write to them (likely 0).

So for your 100 the result would likely be (100 & 0xC7) which happens to be 68.

This problem does not explain however why your get_MSB returns an unexpected 0 instead of a 1.

Make your life easier by not using peripheral registers to check that a function returns the right value. If your debugger is horrible and does not allow to step into a simple

if (100 == get_LSB(100))
{
   volatile unsigned char dummy = 0U;
}

You could instead toggle a pin (which you know you can measure and works correctly).