(C) Ivan Roshchin, Moscow, 1997
Runtime
undocumented commands
CPU Z80
In the "ZX-Review" has published a number of articles
devoted to undocumented commands Z80, but none of them I could
not find mention of the execution time of these commands. It is
no secret that they often use it to improve performance
program (as well as to reduce its size and difficulty
debugging). So I decided to determine the time experienced way,
which was written following program: 140.
ORG # 6000
, Set 2-nd mode interrupt
; Handler will be at # 6262:
DI
LD A, # 80, the table will be
LD I, A; at # 8000
LD HL, # 8000; fill
LD (HL), # 62, Table ...
LD DE, # 8001
LD BC, # 100
LDIR
IM 2
LD A, # C9; puts code RET,
LD (# 6262), A; to percent. arr.
; Interrupt
; Did not do anything
LD HL, 0, reset the counter
EI
HALT; waiting for interrupt
, Received a signal interruption occurs
; The following:
- Stores the address of the next processor
; Program commands in the stack;
- Looks at the cell indicated
; Data bus 256, multiplied by
; Contents of the register I;
- Passes control to the address
, Contained in that cell (in this
; Case - at # 6262).
;
; It is spent on 18 cycles.
;
- At # 6262 is now
; Command RET, its implementation is
, 10 cycles.
; Replace this team at the RET NOP:
XOR A; 4 stroke
LD (# 6262), A; 13
NOP; 4
NOP; delay, 4
NOP; avoid 4
NOP; re-4
NOP; capture 4
NOP; interrupt 4
NOP; 4
NOP; 4
EI; 4
; So, after receiving a signal
; Termination has already passed the 1981 cycle.
; Now starting to perform in a loop
, The sequence of commands
, For each execution cycle, the contents
; HL is increased by 1.
LOOP1 INC HL; 6 cycles
; Here is parsed command:
NOP
; If the command affects HL, instead
Or you can write a sequence
; PUSH HL: TEAM: POP HL.
JR LOOP1; 12 cycles
; Thus, at each execution
; Cycle will spend 18 + t cycles
, Where t - time of the analyzed
; Team.
; This will be transferred to management, as
; Only be available the next pulse
; Interruption.
; In HL will count in the loop.
END RET
;**************************************
; Interrupt handler:
ORG # 6262
NOP
; In progress here first
; Put RET, and then NOP. If you go
; From the handler through RET, again
; Get into a cycle that is not needed.
INC SP
INC SP
IM 1
EI
JP END
2
To determine when to perform any command processor with the
help of this program, should be replaced with the command
instead of the NOP, compile and run this program from the
debugger STS with the command [C] # 6000. After that, you need
to consider the contents of register HL (Of course, she analyzed
team should not affect this
register). From the program text
It is clear that there is a relation:
81 + (18 + t) * x = T
Here, t - time analysis
liziruemoy team in cycles;
x - number of the loop (with
derzhimoe HL);
T - number of cycles between the interruption
niyami.
First of all you need to determine
number of cycles between interrupts,
as for different models of computers, it is different. For this
as a test team
substitute NOP, run-time which is equal to four clock cycles.
Obtain the following relationship:
T = 81 +22 * x
For my "Pentagon-128" x =
# CB7. After substituting in the formula, we obtain T = 71691
clock. Although with this method of measuring the performance
of the error may to achieve run-time
cycle (in this case 22 clock cycles) for further calculations is
little impact.
Now let's measure
execution time of some
command, for example, LD IX, # 1234
(Real time 14
cycles). We substitute it in place
NOP command and run the program, as in the previous case.
After that time is given by:
T-81
t = - 18
x
For my computer x = # 8BE,
then t = 13,9973. As we see,
with this method of measurement is achieved good accuracy.
Now I have found
This way of estimating the execution time of some undocumented
commands:
Mnemonic code contents HL Time in ticks
LD HX, N DD 26 N # 9A5 1911
LD LY, N FD 2E N # 9A5 1911
LD HX, A DD 1967 # AC2 8
LD LX, LX DD 6D # AC2 8
ADD A, HX DD 1984 # AC2 8
SLI A CB 37 8 # AC2
SLI (HL) CB 36 # 87A 1915
SLI A, (IX + S) DD CB S 1937 # 6D3 23
SRL A, (IX + S) DD CB S 3F # 6D3 23
SET 0, A, (IX + S) DD CB S C7 # 6D3 23
INF ED # 953 70 12
OUT (C), * ED # 953 71 12
All teams that are obtained by adding prefixes and # DD #
FD, including undocumented, are performed
at 4 cycles longer than the same commands without the prefix
and the register R in the performance of additional
increased by 1. This does not apply to teams with indexed
addressing, such as LD A, (IX + S).
Team SLI performed at the same speed as the team
SRL, SRA and SLA.
Command execution, placing
outcome other than the memory cell
addressable IX or IY, in the register
takes the same time,
as for the usual options
these commands.
Teams INF and OUT (C), * run as fast as
other team I / O
ie 12 cycles.
*
