C: How do I simulate 8086 registers?

Question

Ohai, I'm currently trying to implement an 8086 ASM debugger for learning purposes. Until now, I tried to simulate the 8 and 16 bit registers with char arrays but this approach is driving me nuts, when working with AX, AL and AH.

#define setAL() { int i; for (i = 0; i < 8; i++) AL[i] = AX[i]; }
char AX[16]   = {0, 1, 1, 1, 1 ,1 ,1, 0, 0, 0, 0, 0, 0, 0, 0, 0};
char AL[8]    = {0, 0, 0, 0, 0, 0, 0, 0};

Does anyone has any good idea (or something like 'best practice') how to simulate those registers?

Answer 1

I don't think there's a 'best practice' way of doing this, but one approach you could take that may drive you nuts less is to use a union to overlay the 8 and 16 bit portions:

struct RegByte { 
   unsigned char low;
   unsigned char high;
};

struct RegWord {
   unsigned short value;
};

union Reg {
   struct RegWord word;
   struct RegByte bytes;
};

Alternatively given you're explicitly targeting just 8086 you could have one structure containing all the 16 bit registers and one containing all of the byte portions. e.g.

struct RegByte {
   unsigned char al, ah, bl, bh, cl, ch, dl, dh;
};

struct RegWord {
   unsigned short ax, bx, cx, dx;
   /* nothing stopping you from continuing with si, di, etc even though
    * they don't have addressable high and low bytes */
};

union Reg {
   struct RegWord word;
   struct RegByte byte;
};

Answer 2

I'd abstract the structure away and use accessor functions.

struct registry_file_t;
uint16_t get_al(registry_file_t * r);
void set_al(registry_file_t * r, uint16_t value);

With inlining enabled, this approach will be no less performant than using a union.