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This page last updated on 28 Aug 1999

[Z80 Tech Info] [48K Tech Info] [128K Tech Info] [Peripheral Ports]

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Most Z80 opcodes are one byte long, not counting a possible byte or word operand. The four opcodes CB, DD, ED and FD are opcodes: they change the meaning of the opcode following them.

CB opcodes

There are 248 different CB opcodes. The block CB 30 to CB 37 is missing from the official list. These instructions, usually denoted by the mnemonic SLL, Shift Left Logical, left the operand and make bit 0 always one. These instructions are quite commonly used. For example, Bounder and Enduro Racer use them.

ED opcodes

There are a number of unofficial ED instructions, but none of them are very useful. The ED opcodes in the range 00-3F and 80-FF (except for the block instructions of course) do nothing at all but taking up 8 T states and incrementing the R register by 2. Most of the unlisted opcodes in the range 40-7F do have an effect, however. The complete list:
(* = not official, i.e. not listed in Zilog documentation)

        ED40   IN B,(C)                 ED60   IN H,(C)
        ED41   OUT (C),B                ED61   OUT (C),H
        ED42   SBC HL,BC                ED62   SBC HL,HL
        ED43   LD (nn),BC               ED63   LD (nn),HL
        ED44   NEG                      ED64 * NEG
        ED45   RETN                     ED65 * RETN
        ED46   IM 0                     ED66 * IM 0
        ED47   LD I,A                   ED67   RRD
        ED48   IN C,(C)                 ED68   IN L,(C)
        ED49   OUT (C),C                ED69   OUT (C),L
        ED4A   ADC HL,BC                ED6A   ADC HL,HL
        ED4B   LD BC,(nn)               ED6B   LD HL,(nn)
        ED4C * NEG                      ED6C * NEG
        ED4D   RETI                     ED6D * RETN
        ED4E * IM 0/1                   ED6E * IM 0/1
        ED4F   LD R,A                   ED6F   RLD
        ED50   IN D,(C)                 ED70   IN (C)
        ED51   OUT (C),D                ED71 * OUT (C),0
        ED52   SBC HL,DE                ED72   SBC HL,SP
        ED53   LD (nn),DE               ED73   LD (nn),SP
        ED54 * NEG                      ED74 * NEG
        ED55 * RETN                     ED75 * RETN
        ED56   IM 1                     ED76 * IM 1
        ED57   LD A,I                   ED77 * NOP
        ED58   IN E,(C)                 ED78   IN A,(C)
        ED59   OUT (C),E                ED79   OUT (C),A
        ED5A   ADC HL,DE                ED7A   ADC HL,SP
        ED5B   LD DE,(nn)               ED7B   LD SP,(nn)
        ED5C * NEG                      ED7C * NEG
        ED5D * RETN                     ED7D * RETN
        ED5E   IM 2                     ED7E * IM 2
        ED5F   LD A,R                   ED7F * NOP

The ED70 instruction reads from port (C), just like the other instructions, but throws away the result. It does change the flags in the same way as the other IN instructions, however. The ED71 instruction OUTs a zero byte to port (C). These instructions 'should', by regularity of the instruction set, use (HL) as operand, but since from the processor's point of view accessing memory or accessing I/O devices is the same thing except for activation of the /IORQ line instead of the /MREQ line, and since the Z80 cannot access memory twice in one instruction (disregarding instruction fetch of course), it can't fetch or store the data byte.

The instructions ED 4E and ED 6E are IM 0 equivalents: when FF was put on the bus (physically) at interrupt time, the Spectrum continued to execute normally, whereas when an EF (RST #28) was put on the bus it crashed, just as it does in that case when the Z80 is in the official interrupt mode 0. In IM 1 the Z80 just executes a RST #38 (opcode FF) no matter what is on the bus.

All the EDxx RET? instructions copy IFF2 to IFF1, even RETI (ED4D), which the official documentation does not note. The only difference between RETI and RETN is that peripheral devices which allow daisy-chaining of interrupts (eg the Z80 PIO) recognise the ED4D sequence as 'end of interrupt' and then know that they can allow a further interrupt to be passed to the processor.

DD and FD opcodes

The DD and FD opcodes precede instructions using the IX and IY registers. If you look at the instructions carefully, you see how they work:

    2A nn      LD HL,(nn)
    DD 2A nn   LD IX,(nn)
    7E         LD A,(HL)
    DD 7E d    LD A,(IX+d)

A DD opcode simply changes the meaning of HL in the next instruction. If a memory byte is addressed indirectly via HL, as in the second example, a displacement byte is added. Otherwise the instruction simply acts on IX instead of HL (A notational awkwardness, that will only bother assembler and disassembler writers: JP (HL) is not indirect; it should have been denoted by JP HL). Instructions which use H or L access the high and low halves of IX; those which reference both (HL) and either H or L replace HL by (IX+d), but still use H or L. For example, DD6601 is LD H,(IX+01). Very many programs use these 'undocumented' IX instructions. FD works in exactly the same way to DD, but with IY instead of IX. Many DD or FD opcodes after each other will effectively be NOPs, doing nothing except repeatedly setting the flag "treat HL as IX" (or IY) and taking up 4 T states (But try to let MONS disassemble such a block.).

It is also possible to have doubly-shifted DDCB and FDCB opcodes (if DD or FD precedes an ED instruction, the DD or FD is ignored, meaning that the ED instructions never operate on IX or IY). With the CB instructions, the situation is more interesting. Every DDCB instruction operates on (IX+nn), but also copies the result to the register used in the original instruction, except when it is (HL). For example,

        CB CE           SET 0,(HL)
        CB C0           SET 0,B
        DD CB nn CE     SET 0,(IX+nn)
        DD CB nn C0     SET 0,(IX+nn) ; copy result to B

There is no standard way to denote these doubly shifted opcodes. Also, note that the offset byte is the third under all circumstances: for the singly shifted opcodes, the offset byte is after the opcode byte (eg DD 2A nn is LD HL,(nn)), whilst the doubly shifted opcodes have the offset byte before the opcode byte.

The R Register

This is not really an undocumented feature, but a thorough description of it is not easy to find. The R register is a counter that is updated during every Z80 M1 cycle (approximately equivalent to every instruction), so long as DD, FD, ED and CB are to be regarded as separate instructions, so shifted instructions increase R by two. There's an interesting exception: doubly-shifted opcodes, the DDCB and FDCB ones, also increase R by two. LDI increases R by two, LDIR increases it by 2 times BC, as does LDDR etcetera. R is set to zero when the Z80 is reset.

Both LD A,R and LD R,A use the value of R after it has been increased (eg an XOR A/LD R,A sequence sets the value of R to zero, and [reset]/DI/LD A,R sets A to #03).

The highest bit of the R register is never changed: this is because in the old days everyone used 16 Kbit chips. Inside the chip the bits where grouped in a 128x128 matrix, needing a 7 bit refresh cycle. Therefore ZiLOG decided to count only the lowest 7 bits. You can easily check that the R register is really crucial to memory refresh. Assemble this program:

        ORG 32768
        LD B,0
    L1: XOR A
        LD R,A
        DEC HL
        LD A,H
        OR L
        JR NZ,L1
        DJNZ L1

It will take about three minutes to run. Look at the upper 32K of memory, for instance the UDG graphics. It will have faded. Only the first few bytes of each 256 byte block will still contain zeros, because they were refreshed during the execution of the loop. The ULA took care of the refreshing of the lower 16K (This example won't work on the emulator, of course!).

R is increased by 1 during interrupt or NMI acknowledge (ignore the complicated, incorrect info which was in previous versions of this FAQ).

Undocumented Flags

This undocumented "feature" of Z80 has its effect on programs like Sabre Wulf, Ghosts'n'Goblins and the Speedlock loaders. Bits 3 and 5 of the F register are not used. They can contain information, as you can readily figure out by PUSHing AF onto the stack and then POPping some it into another pair of registers. Furthermore, sometimes their values change. The following empirical rule (due to Gerton Lunter) gives their values after most instructions:

The values of bits 5 and 3 follow the values of the corresponding bits of the last 8 bit result of an instruction that changed the usual flags.

For instance, after an ADD A,B those bits will be identical to the bits of the A register.

As well as the two completely undocumented flags, after some instructions, the official documentation lists the value of some flags as 'undefined'. However, these flags have predictable values: (In the list below, C is the register and c is the carry flag)

     Instruction                        Non-standard flags

CP xx                     3 and 5 copied from the argument, not the result

ADD HL,xx                 Consider the instruction being done in two steps:
ADC HL,xx/SBC HL,xx       first the LSBs being added, then the MSBs. The
                          3,H,5 and S flags are set as for the second step,
                          and Z is set only if the entire 16-bit result is
                          zero. (S and Z are not changed by ADD HL,xx).

BIT n,r			  P/V is set to the same value as Z. S is reset
                          unless the instruction is BIT 7,r and bit 7
                          of r is set, in which case S is set.

BIT n,(HL)                3 and 5 are apparently copied from an internal
BIT n,(IX/IY+d)           storage in the Z80; this is set as follows:
                          ADD HL,xx: H before the addition
                          LD r,(IX/IY+d): high byte of IX/IY+d
                          JR d: high byte of the jump target
                          LD r,r': no effect
                          Others have not been tested yet.

SCF/CCF/CPL               3 and 5 copied from A. CCF sets H to the value of
                          c before the instruction is executed.

LDD/LDDR/LDI/LDIR         3 is bit 3 of (copied value+A), whilst 5 is bit 1 of
                          this value.

CPD/CPDR/CPI/CPIR	  3 is bit 3 of (A-(HL)-(half carry flag)); 5 is bit 1
                          of this value. (HL) is the value of (HL) before the
                          instruction, whilst H is the value of H after the

IND/INDR/INI/INIR         S,5 and 3 are affected as DEC B; N is set to bit
OUTD/OTDR/OUTI/OTIR       7 of the value written to/read from the IO port. c
                          is found by taking C, adding one if the instruction
                          increments HL or decrementing it otherwise, then
                          adding the value written/read, and taking the carry
                          of this final sum. H is set to the same value as c.

P/V for the IN... and OUT... instructions can be calculated as follows: in the following, x.y refers to bit y of x and inp is the byte read from the port. Look at bits 0 and 1 of C and inp, and obtain a temporary result. The first result column should be used for IND/INDR/OUTD & OTDR and the second for INI/INIR/OUTI & OTIR:

    C.1   C.0  inp.1 inp.0   Temp1
     0     0     0     0      0/0
     0     0     0     1      1/0
     0     0     1     0      0/1
     0     0     1     1      0/0
     0     1     0     0      1/0
     0     1     0     1      0/1
     0     1     1     0      0/0
     0     1     1     1      1/1
     1     0     0     0      0/1
     1     0     0     1      0/0
     1     0     1     0      1/1
     1     0     1     1      0/1
     1     1     0     0      0/0
     1     1     0     1      1/1
     1     1     1     0      0/1
     1     1     1     1      1/0

Now, calculate Temp2 according to the following pseudo-code:

If B.3 == B.2 == B.1 == B.0 == 0 then
  Temp2 = Parity(B) xor (B.4 or (B.6 and not B.5))
  Temp2 = Parity(B) xor (B.0 or (B.2 and not B.1))
(Parity(B) is the standard partity function). Finally,
P/V = Temp1 xor Temp2 xor C.2 xor inp.2

Ghosts'n'Goblins uses the undocumented flag due to a programming error. The rhino in Sabre Wulf walks backward or keeps running in little circles in a corner, if the (in this case undocumented) behaviour of the sign flag in the BIT instruction isn't right. From the code:

        AD86    DD CB 06 7E        BIT 7,(IX+6)
        AD8A    F2 8F AD           JP P,#AD8F

An amazing piece of code! Speedlock does so many weird things that all must be exactly right for it to run. Finally, the '128 ROM uses the AF register to hold the return address of a subroutine for a while.


The Z80 has three interrupt modes, selected by the instructions IM 0, IM 1 and IM 2.

When an interrupt is due, which is signalled by the ULA taking the level-triggered /INT pin on the Z80 low, nothing happens until the last M-cycle of the instruction currently being executed. At that point, if interrupts are enabled (IFF1 is set) then interrupt processing will begin. For this purpose, HALT is effectively an infinite series of NOPs, and the repeated instructions (LDIR, etc) can be interrupted after each execution. Interrupt processing begins by resetting IFF1 and IFF2; this has two non-obvious consequences:

The flip flops IFF1 and IFF2 are simultaneously set or reset by the EI and DI instructions. IFF1 determines whether interrupts are allowed, but its value cannot be read; IFF2 can be copied to the P/V flag as mentioned above.

On the 48K Spectrum, the ULA holds the /INT pin low for precisely 32 T-states. This pin is sampled during the last M-cycle of every instruction apart from repeated IX and IY prefixes (DD and FD). If the pin goes high again before it is sampled, no interrupt will occur. The /INT pin must be held low for at least 23 T-states, as some IX and IY instructions take 23 T-states. If the interrupt routine starts EI/NOP, this can cause a double interrupt, as the /INT pin will be sampled 19 (for IM 2)+4+4=27 T-states after being first sampled, when it may still be low.

In IM 1, the processor simply executes an RST #38 instruction if an interrupt is requested. This is the mode the Spectrum is initalised to. In this mode, the processor takes 13 T states to reach #0038: a 7 T state M1 cycle to acknowledge the interrupt and decrement SP, a 3 T state M2 cycle to write the high byte of PC onto the stack and decrement SP again, and finally a 3 T state M3 cycle to write the low byte onto the stack and to set PC to #0038.

The other mode that is commonly used on the Spectrum is IM 2. If an interrupt is requested, the processor first builds a 16 bit address by combining the I register (as the high byte) with whatever the interrupting device places on the data bus. The processor then fetches the 16-bit address at this interrupt table entry, and finally calls the subroutine at that address. Rodnay Zaks in his book 'Programming the Z80' states that only even bytes are allowed as low index byte, but that isn't true. The normal Spectrum contains no hardware to place a byte on the bus, and the bus will therefore always read FF (because the ULA also doesn't read the screen if it generates an interrupt), so the resulting index address is 256*I+255. However, some not-so-neat hardware devices put things on the data bus when they shouldn't, so later programs didn't assume the low index byte was FF. These programs contain a 257 byte table of equal bytes starting at 256*I, and the interrupt routine is placed at an address that is a multiple of 257. A useful but not so much used trick on the Spectrum is to make the table contain FF's (or use the ROM for this) and put a byte 18 hex, the opcode for JR, at FFFF. The first byte of the ROM is a DI, F3 hex, so the JR will jump to FFF4, where a long JP to the actual interrupt routine is put. In IM 2, it takes 19 cycles to get to the interrupt routine:

In interrupt mode 0, the processor executes the instruction that the interrupting device places on the data bus. On a standard Spectrum this will be the byte FF, coincidentally (:-)) the opcode for RST #38. But for the same reasons as above, this is not really reliable. If there is a RST n on the data bus, it takes 12 cycles to get to 'n':

With a CALL nnnn on the data bus, it takes 19 cycles:

When the /NMI pin goes low, an internal flip-flop in the Z80 is set to note that an NMI is pending; This flip-flop is sampled at the end of every instruction, apart from DD/FD and possibly EI/DI.

When an NMI occurs, IFF1 is reset, thereby disallowing further maskable interrupts, but IFF2 is left unchanged. This enables the NMI service routine to check whether the interrupted program had enabled or disabled maskable interrupts. The NMI routine should end with a RETN instruction, which in addition to the usual RET actions copies IFF2 to IFF1, thus restoring the interrupt state of the interrupted code.

When an NMI occurs, it takes 11 T states to get to #0066: a 5 T state M1 cycle to do an opcode read and decrement SP, a 3 T state M2 cycle to write the high byte of PC to the stack and decrement SP and finally a 3 T state M3 cycle to write the low byte of PC and jump to #0066.

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This FAQ is maintained by Philip Kendall; distribution is permitted only under the conditions specified in the copyright notice.
Primary site for this FAQ: http://www.ast.cam.ac.uk/~pak/cssfaq/index.html.