cemulationadditionx86-1616-bit

function that adds 16-bit variables

I am writing my first emulator (Intel 8086 microprocessor).

I'm stuck because I have to write a function that adds 16-bit variables together. Here is the prototype of the function I want to write:

uint16_t add(uint16_t reg, uint16_t operand);

For example, if we execute :

add(5, 3)

The function should do the calculation like this:

              111
  0000000000000101 (= 5)
+ 0000000000000011 (= 3)
  ----------------
              1000 (= 8)

and return 8.

To write this function, I need to find a way to access each bit of the two variables, then to be able to add them together and place them in a third variable.

Is this possible in C and if so can someone enlighten me on this?

Thank you very much for your answers.

Solution

  • Assuming 32 bit computer, this is very trivial:

    #include <stdint.h>
#include <stdio.h>

_Bool carry;

uint16_t add (uint16_t reg, uint16_t operand)
{
  carry = (reg + operand) & 0xFFFF0000u;
  return reg + operand;
}

int main (void)
{
  int sum = add(65535, 1);
  printf("%d, CF:%d\n", sum, (int)carry);
}

    Where reg + operand will get promoted to 32 bit int and in case the addition goes beyond 16 bits, the boolean flag will get set to 1.