Question about "Immediate Data" in Assembly

Question

Hi, In assembly, when they say "immediate data" is that signed or unsigned??

I'm writing a Gameboy emulator and am using the opcodes here:

http://www.pastraiser.com/cpu/gameboy/gameboy_opcodes.html

Opcode 0xC6 for example is ADD A, d8.

My guess is that it's unsigned else why would they need "SUB A, d8" but I thought that I'd ask just while I'm checking over my code...

Thanks in advance, Keith

Answer 1

It seems like it's an unsigned int. Look at MAME's LR35902 CPU emulator, they use UINT8 for the immediate.

Some relevant code:

/* ... */
case 0xC6: /*      ADD A,n8 */
    x = mem_ReadByte (cpustate, cpustate->w.PC++);
    ADD_A_X (x)
    break;
/* ... */

#define ADD_A_X(x) \
     { \
       register UINT16 r1,r2; \
       register UINT8 f; \
       r1=(UINT16)((cpustate->b.A&0xF)+((x)&0xF)); \
       r2=(UINT16)(cpustate->b.A+(x)); \
       cpustate->b.A=(UINT8)r2; \
       if( ((UINT8)r2)==0 ) f=FLAG_Z; \
         else f=0; \
       if( r2>0xFF ) f|=FLAG_C; \
       if( r1>0xF )  f|=FLAG_H; \
       cpustate->b.F=f; \
     }

Now, the routine for LD HL,SP+n8 is

case 0xF8: /*      LD HL,SP+n8 */
/*
 *   n = one UINT8 signed immediate value.
 */

  {
    register INT32 n;
    n = (INT32) ((INT8) mem_ReadByte (cpustate, cpustate->w.PC++));

Here the immediate is cast to an signed 8-bit int (INT8), so I imagine it's otherwise unsigned.

Answer 2

In a two's compliment system, it doesn't actually matter of it is signed or unsigned.

Interpretation of the value (and overflow) happens when you do a comparison instruction. Until that point it is just bits. Adding 0x8000 to 0x0001 gives 0x8001 whether or not either of the values is signed or unsigned.

You should look at the opcodes for ADD and SUB. It could be that there are not actually two different operations. It could be that SUB just negates the operand and then issues an ADD instruction.