Z80 DAA instruction

Question

Apologies for this seemingly minor question, but I can\'t seem to find the answer anywhere - I\'m just coming up to implementing the DAA instruction in my Z80 emulator, and I no

Answer 1

Does it just modify the accumulator anyway, based on the conditions set out in the DAA table, regardless of the previous instruction?

Yes. The documentation is only telling you what DAA is intended to be used for. Perhaps you are referring to the table at this link:

--------------------------------------------------------------------------------
|           | C Flag  | HEX value in | H Flag | HEX value in | Number  | C flag|
| Operation | Before  | upper digit  | Before | lower digit  | added   | After |
|           | DAA     | (bit 7-4)    | DAA    | (bit 3-0)    | to byte | DAA   |
|------------------------------------------------------------------------------|
|           |    0    |     0-9      |   0    |     0-9      |   00    |   0   |
|   ADD     |    0    |     0-8      |   0    |     A-F      |   06    |   0   |
|           |    0    |     0-9      |   1    |     0-3      |   06    |   0   |
|   ADC     |    0    |     A-F      |   0    |     0-9      |   60    |   1   |
|           |    0    |     9-F      |   0    |     A-F      |   66    |   1   |
|   INC     |    0    |     A-F      |   1    |     0-3      |   66    |   1   |
|           |    1    |     0-2      |   0    |     0-9      |   60    |   1   |
|           |    1    |     0-2      |   0    |     A-F      |   66    |   1   |
|           |    1    |     0-3      |   1    |     0-3      |   66    |   1   |
|------------------------------------------------------------------------------|
|   SUB     |    0    |     0-9      |   0    |     0-9      |   00    |   0   |
|   SBC     |    0    |     0-8      |   1    |     6-F      |   FA    |   0   |
|   DEC     |    1    |     7-F      |   0    |     0-9      |   A0    |   1   |
|   NEG     |    1    |     6-F      |   1    |     6-F      |   9A    |   1   |
|------------------------------------------------------------------------------|

I must say, I've never seen a dafter instruction spec. If you examine the table carefully, you will see that the effect of the instruction depends only on the C and H flags and the value in the accumulator -- it doesn't depend on the previous instruction at all. Also, it doesn't divulge what happens if, for example, C=0, H=1, and the lower digit in the accumulator is 4 or 5. So you will have to execute a NOP in such cases, or generate an error message, or something.

Answer 2

Just wanted to add that the N flag is what they mean when they talk about the previous operation. Additions set N = 0, subtractions set N = 1. Thus the contents of the A register and the C, H and N flags determine the result.

The instruction is intended to support BCD arithmetic but has other uses. Consider this code:

    and  15
    add  a,90h
    daa
    adc  a,40h
    daa

It ends converting the lower 4 bits of A register into the ASCII values '0', '1', ... '9', 'A', 'B', ..., 'F'. In other words, a binary to hexadecimal converter.

Answer 3

I found this instruction rather confusing as well, but I found this description of its behavior from z80-heaven to be most helpful.

When this instruction is executed, the A register is BCD corrected using the contents of the flags. The exact process is the following: if the least significant four bits of A contain a non-BCD digit (i. e. it is greater than 9) or the H flag is set, then $06 is added to the register. Then the four most significant bits are checked. If this more significant digit also happens to be greater than 9 or the C flag is set, then $60 is added.

This provides a simple pattern for the instruction:

• if the lower 4 bits form a number greater than 9 or H is set, add $06 to the accumulator
• if the upper 4 bits form a number greater than 9 or C is set, add $60 to the accumulator

Also, while DAA is intended to be run after an addition or subtraction, it can be run at any time.

Answer 4

This is code in production, implementing DAA correctly and passes the zexall/zexdoc/z80test Z80 opcode test suits.

Based on The Undocumented Z80 Documented, pag 17-18.

void daa()
{
   int t;
    
   t=0;
    
   // 4 T states
   T(4);
    
   if(flags.H || ((A & 0xF) > 9) )
         t++;
    
   if(flags.C || (A > 0x99) )
   {
         t += 2;
         flags.C = 1;
   }
    
   // builds final H flag
   if (flags.N && !flags.H)
      flags.H=0;
   else
   {
       if (flags.N && flags.H)
          flags.H = (((A & 0x0F)) < 6);
       else
          flags.H = ((A & 0x0F) >= 0x0A);
   }
    
   switch(t)
   {
        case 1:
            A += (flags.N)?0xFA:0x06; // -6:6
            break;
        case 2:
            A += (flags.N)?0xA0:0x60; // -0x60:0x60
            break;
        case 3:
            A += (flags.N)?0x9A:0x66; // -0x66:0x66
            break;
   }
    
   flags.S = (A & BIT_7);
   flags.Z = !A;
   flags.P = parity(A);
   flags.X = A & BIT_5;
   flags.Y = A & BIT_3;
}

For visualising the DAA interactions, for debugging purposes, I have written a small Z80 assembly program, that can be run in an actual ZX Spectrum or in an emulation that emulates accurately DAA: https://github.com/ruyrybeyro/daatable

As how it behaves, got a table of flags N,C,H and register A before and after DAA produced with the aforementioned assembly program: https://github.com/ruyrybeyro/daatable/blob/master/daaoutput.txt