更改程序集8086中寄存器的符号位

Question

I'm trying to write assembly code that does a fairly simple thing, it changes the sign bit (just the sign bit) of the register AL.

I need to find two different ways to do this, but sadly the solutions I came up with don't seem to be working.

1) I tried subtracting AL from itself twice:

mov BL,AL
sub AL,BL
sub AL,BL

But this changes more than just the sign bit. And I understand why.

2) I also tried adding 10000000 to AL. This indeed solves the problem, but if the sign bit of AL is 1, then I get an overflow and carry.

Is there a way to change just the sign bit, without running into overflow / carry problems?

Answer 1

How about just doing:

XOR AL, 80H

?

also:

MOV BL, AL
MOV AL, 0
SUB AL, BL
AND BL, 7FH
AND AL, BL

Another thing to preserve the flags you can always go

   PUSHF
       // whatever
    POPF

Answer 2

Shift it into the carry flag, complement the flag and shift it back:

shl al, 1
cmc
rcr al, 1

To preserve the flags you can use pushf and popf .

Answer 3

XOR operation the constant 0x8000 (for 16bit) with the register you want to change.

Bit of googling:

xor Bitwise logical XOR
  Syntax:   xor dest, src
  dest: register or memory
  src: register, memory, or immediate
  Action: dest = dest ^ src
  Flags Affected: OF=0, SF, ZF, AF=?, PF, CF=0

Answer 4

To change sign use NEG instruction

for example

MOV AX,0101        - AX = 0101
NEG AX             - AX = 1010
                      because of the NEG instruction


NEG instruction affects these flags only:
    CF, ZF, SF, OF, PF, AF.

NEG - Make operand negative (two's complement).
      Actually it reverses each bit of operand and then adds 1 to
      it. For example 5 will become -5, and -2 will become 2.