Question

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…

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.