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Demos Amiga Demoscene Archive Forum / Coding / Calculating time and number of cycles

 

Author Message
nemas
Member
#1 - Posted: 22 Sep 2012 17:20
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Hi,

I want to try to calculate how much time (and cycles) a routine takes. This is the code I have implemented but I'm seeing some odd behaviour with it (sorry if the code is a bit messy but mostly for testing and just figuring things out)


include "exec/types.i"
include "exec/io.i"
include "lvo/exec_lib.i"
include "lvo/dos_lib.i"
include "lvo/timer_lib.i"

section code, code

start
lea _dosname,a1
move.l $4.w,a6
jsr _LVOOpenLibrary(a6)
tst.l d0
beq exit

move.l d0,dos_handle

; ==============================
; Open timer.device

lea timer_device_name,a0
lea timer_device,a1
moveq #0,d0
moveq #0,d1
jsr _LVOOpenDevice(a6)
tst.l d0
bne .end

lea timer_device,a1
move.l IO_DEVICE(a1),a6

moveq #8-1,d4

.calc_loop

lea time_output,a0
jsr _LVOReadEClock(a6)

; Code to time goes here...

move.w #40000,d2
moveq #1,d3
.test_loop
nop
dbra d2,.test_loop

; Code to time ends here...

lea time_output2,a0
jsr _LVOReadEClock(a6)

lea format_data,a1
lea time_output,a0
move.l 4(a0),(a1)
jsr print_timing

lea format_data,a1
lea time_output2,a2
move.l 4(a2),(a1)
jsr print_timing

lea time_output,a0
lea format_data,a1
lea time_output2,a2
move.l 4(a2),d0
sub.l 4(a0),d0
move.l d0,(a1)
jsr print_timing

dbra d4,.calc_loop

lea timer_device,a1
move.l $4.w,a6
jsr _LVOCloseDevice(a6)

.end
move.l dos_handle,a1
move.l $4.w,a6
jsr _LVOCloseLibrary(a6)
exit:
rts

print_timing
move.l dos_handle,a6
move.l #format_msg,d1
move.l #format_data,d2
jsr _LVOVPrintf(a6)
rts

section data, data

format_msg dc.b 'time %ld\n',0
align 4
format_data dc.l 10

timer_device_name dc.b 'timer.device', 0
_dosname dc.b "dos.library",0

section bss, bss

time_output ds.b 8
time_output2 ds.b 8
timer_device ds.b 512
dos_handle ds.b 4


So the way this code works is quite straightforward as it opens dos.library (for printing stuff) and then timer.device and user ReadEClock for timing.

The routine is looped 8 times to mesure differences in the timing. And then grabs the lower 32-bit value from the timing output (which is in microseconds afaik) so far so good.

Now if I run the code I get:


time 1099754360
time 1099760305
time 5945
time 1099754360
time 1099760305
time 5945
time 1099754360
time 1099760305
time 5945
time 1099754360
time 1099760305
time 5945
...


Still looks good but if I run it again:


time 1342377417
time 1342384176
time 6759
time 1342377417
time 1342384176
time 6759
time 1342377417
time 1342384176
time 6759
time 1342377417
time 1342384176
time 6759
...


so the time here is quite different from before and the timing is consistent also. So my questions:

1. Is the timer.device/ReadEClock the best way to calculate the time?
2. What is the cause of the unbalanced timing? (is it caused by the OS still running and might "interfere" with the timing or is the timer just crap? :)
3. When I got the actual time (in microseconds) how do I calculate an ~approx cycle count for the routine given a specific MHz that is (say 060/55Mhz)

Thanks for any help,

/Nemas
hitchhikr
Member
#2 - Posted: 23 Sep 2012 19:40
Reply Quote
You'll have to disable the interruptions when running your code otherwise they will interfere, cache memory work may also modify results.
nemas
Member
#3 - Posted: 23 Sep 2012 20:21
Reply Quote
Hi,

I tried it now and it works much better it can still diff a bit but its way better so it should be fine (note to self also: nop on 060 is really slow (not that you really use it but still))

So that solves 2. Anyone knows about 1, 3.

Thanks!

/Nemas

 

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