On my machine - I did 4 cases of dividing 40 by 2 and the slowest was "x =
BITRSHIFT(40,2)"
Based on 1,000,000 iterations for each I get the following timings ...
x = 40 / 4 .. 0.1070546391
STORE 40 / 4 TO x .. 0.0636483880
x = BITRSHIFT(40,2) .. 0.1302353828
STORE BITRSHIFT(40,2) TO x .. 0.0844159499
I don't think the difference is too dramatic.
If I change the 40 to 51.2 the figures are slightly higher but in same
proportions to each other | x = BITRSHIFT(51.2,2) is still the slowest (and
of course inaccurate)
Changing to 2E9 I get pretty much the same timings as the first set.
Code I used .....
ACTIVATE SCREEN
CLEAR
SET DECIMALS TO 8
DECLARE INTEGER QueryPerformanceCounter IN C:\Windows\SysWow64\kernel32.dll;
STRING @lpPerformanceCount
DECLARE INTEGER QueryPerformanceFrequency IN
C:\Windows\SysWow64\kernel32.dll;
STRING @lpFrequency
*- Get Timer Resolution / Frequency
lcBuffer = REPLICATE(CHR(0), 8)
QueryPerformanceFrequency(@lcBuffer)
lnFrequency = buf2dword(SUBSTR(m.lcBuffer, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer, 5, 4)) * 4294967296
*----------------------------------------------------
*- Case 1 | x = 2E9 / 4
*----------------------------------------------------
STORE REPLICATE(CHR(0), 8) TO lcBuffer1, lcBuffer2
QueryPerformanceCounter(@lcBuffer1)
FOR n = 1 TO 1000000
x = 2E9 / 4
ENDFOR
QueryPerformanceCounter(@lcBuffer2)
lnStart = (buf2dword(SUBSTR(m.lcBuffer1, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer1, 5, 4)) * 4294967296) / m.lnFrequency
lnEnd = (buf2dword(SUBSTR(m.lcBuffer2, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer2, 5, 4)) * 4294967296) / m.lnFrequency
?x, lnEnd - lnStart
*----------------------------------------------------
*- Case 2 | STORE 2E9 / 4 TO x
*----------------------------------------------------
STORE REPLICATE(CHR(0), 8) TO lcBuffer1, lcBuffer2
QueryPerformanceCounter(@lcBuffer1)
FOR n = 1 TO 1000000
STORE 2E9 / 4 TO x
ENDFOR
QueryPerformanceCounter(@lcBuffer2)
lnStart = (buf2dword(SUBSTR(m.lcBuffer1, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer1, 5, 4)) * 4294967296) / m.lnFrequency
lnEnd = (buf2dword(SUBSTR(m.lcBuffer2, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer2, 5, 4)) * 4294967296) / m.lnFrequency
?x, lnEnd - lnStart
*----------------------------------------------------
*- Case 3 | x = BITRSHIFT(2E9,2)
*----------------------------------------------------
STORE REPLICATE(CHR(0), 8) TO lcBuffer1, lcBuffer2
QueryPerformanceCounter(@lcBuffer1)
FOR n = 1 TO 1000000
x = BITRSHIFT(2E9,2)
ENDFOR
QueryPerformanceCounter(@lcBuffer2)
lnStart = (buf2dword(SUBSTR(m.lcBuffer1, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer1, 5, 4)) * 4294967296) / m.lnFrequency
lnEnd = (buf2dword(SUBSTR(m.lcBuffer2, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer2, 5, 4)) * 4294967296) / m.lnFrequency
?x, lnEnd - lnStart
*----------------------------------------------------
*- Case 4 | STORE BITRSHIFT(2E9,2) TO x
*----------------------------------------------------
STORE REPLICATE(CHR(0), 8) TO lcBuffer1, lcBuffer2
QueryPerformanceCounter(@lcBuffer1)
FOR n = 1 TO 1000000
STORE BITRSHIFT(2E9,2) TO x
ENDFOR
QueryPerformanceCounter(@lcBuffer2)
lnStart = (buf2dword(SUBSTR(m.lcBuffer1, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer1, 5, 4)) * 4294967296) / m.lnFrequency
lnEnd = (buf2dword(SUBSTR(m.lcBuffer2, 1, 4)) +
buf2dword(SUBSTR(m.lcBuffer2, 5, 4)) * 4294967296) / m.lnFrequency
?x, lnEnd - lnStart
_cliptext = transform(lnEnd - lnstart)
FUNCTION buf2dword (vcBuffer)
RETURN ASC(SUBSTR(m.vcBuffer, 1,1)) + ;
BITLSHIFT(ASC(SUBSTR(m.vcBuffer, 2,1)), 8) +;
BITLSHIFT(ASC(SUBSTR(m.vcBuffer, 3,1)), 16) +;
BITLSHIFT(ASC(SUBSTR(m.vcBuffer, 4,1)), 24)
ENDFUNC
-----Original Message-----
From: ProfoxTech [mailto:[email protected]] On Behalf Of Laurie
Alvey
Sent: Thursday, 30 July 2015 7:58 PM
To: [email protected]
Subject: Re: BITRSHIFT(n,1) v n/2
Interesting. If I issue ? CAST(8E9 As I) it displays a negative 9 digit
number (-589934592). Also, I thought the limit for 32 bit integers was about
2E9.
Laurie
On 29 July 2015 at 13:42, Jean MAURICE <[email protected]> wrote:
> Hi Laurie,
> I do think that BITRSHIFT is a lot quicker than /2. BITRSHIFT can be
> done with only one machine cycle (clock tic); the division is a lot more
complex.
>
> BUT !
> this is true only if K is an integer (k = CAST(8E9 AS integer)).
> Otherwise you must add add the conversion time between numeric and
integer.
>
> You can also add a third test : k * .5 sould be faster than k /2
>
> The Foxil
> Old enough to know how microprocessors work !
>
>
[excessive quoting removed by server]
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