21
votes

I have the next code:

mov al, -5
add al, 132
add al, 1

As I check it, the overflow flag and the carry flag will set in the first operation, and in the second, only the overflow will set.

But I don't understand why:

  1. In unsigned number, the result is 143 (8FH), and for that is fit 8-bit unsigned number (is smaller than 255) => the carry flag shouldn't be set. In signed number, the result is 127, It's fit to 8-bit signed, and the overflow shouldn't be set.

Whats wrong? Thanks.

2
See Understanding Carry vs. Overflow conditions/flags, teaching.idallen.com/dat2343/10f/notes/040_overflow.txtPeter Cordes

2 Answers

19
votes

In unsigned arithmetic, you have added 0xFB to 0x84, i.e. 251 + 132, which indeed is larger than 8-bit, and so the carry flag is set.

In the second case, you are adding +127 to 1, which indeed exceeds a signed 8-bit range, and so the overflow flag is set.

31
votes

Overflow occurs when the result of adding two positive numbers is negative or the result of adding two negative numbers is positive. For instance: +127+1=?

+127=0111 1111
  +1=0000 0001
     ---------
     1000 0000 

As we look at the sign bits of the two operands and the sign bit of the result, we find out that Overflow occurred and the answer is incorrect.