Metastock bug?

jmmac · #1 25th April 2007, 03:29 AM

Hi,

I'm trying something with metastock and the results are not what i was expecting. If you create an indicator with the following code:

a:=0;
a:= Ref(a,-1) + 1;
a;

You get a straight line with the value 1, and i was expecting a line with increasing values, +1 for each new bar. Do you know why this doesn't happen?

I'm using MS Pro 8.0.

Thank you,
Jaime.

oxusmorouz · #2 25th April 2007, 12:50 PM

Quote:

Originally Posted by jmmac

Hi,

I'm trying something with metastock and the results are not what i was expecting. If you create an indicator with the following code:

a:=0;
a:= Ref(a,-1) + 1;
a;

You get a straight line with the value 1, and i was expecting a line with increasing values, +1 for each new bar. Do you know why this doesn't happen?

I'm using MS Pro 8.0.

Thank you,
Jaime.

Hi Jamie,
For continuous adding, why don't you use "cum" function instead?

a:= Cum(1);
a;

As far as I can see, metastock does not recognize 2 variables with the same name and I guess is confused it,lol.

Oxymoron.

jmmac · #3 25th April 2007, 02:55 PM

Hi Oxymoron, thanks for the reply.

My goal isn't really continuous adding, i just simplified my script to post the issue, i cannot use Cum()...

Jaime

oxusmorouz · #4 25th April 2007, 03:03 PM

Quote:

Originally Posted by jmmac

Hi Oxymoron, thanks for the reply.

My goal isn't really continuous adding, i just simplified my script to post the issue, i cannot use Cum()...

Jaime

Your formula:
a:=0;
a:= Ref(a,-1) + 1;
a;

When metastock reads this formula, I guess it uses the previous value of the 1st "a" variable in the reference function which remains constant at 0. This could explain why there is an error.
Try contacting Equis. Maybe they have a solution.

moneypick · #5 25th April 2007, 09:29 PM

Hi jmmac,

Post your problem in the Equis forum-

http://forum.equis.com/forums/default.aspx

uasish · #6 25th April 2007, 10:22 PM

Quote:

Originally Posted by jmmac

Hi Oxymoron, thanks for the reply.

My goal isn't really continuous adding, i just simplified my script to post the issue, i cannot use Cum()...

Jaime

Jaime,

Until you clarify ( we are least interested in your code ,writing MS code for others for money is my past time ,but here in traderji,no question of money rather i have freely given my own codes here) ,all this to alley your worry,we will be unable to help you.

Asish

jmmac · #7 26th April 2007, 10:39 PM

Oxymoron,

Its something like that, yes. But it's not the first value of “a” that is used. If you try this code:

a:=0;
a:=1;
a:=2;
a:= Ref(a,-1) + 1;
a;

MS will plot 3. MS Mysteries!

moneypick,
Maybe i'll try that forum even if it seem a little abandoned.
Thanks.

Asish,
Thank you for the reply.
I can tell you that i tried to translate a small program from wealth-lab developer to MS (for others benefit) and i was unable to do it. It seemed i had found a way but because of the issue mentioned, my idea isn't valid.
I need to refer to the past value of a variable and i cannot for reasons i don’t understand. I don't understand why the code i showed doesn't plot an increasing line, why ref(a,-1) shows always the constant i first assign and not the value i assign later.
I have to assign a constant value first to initialize the vars, otherwise they remain forever NA.

Maybe version 9 or 10 behave in a different manner with this code?

Thank you and regards,
Jaime

jmmac · #8 27th April 2007, 12:23 AM

I just read a message regading this issue in a different forum that may explain MS behaviour in this case. It seems that MS takes each line of script at a time and calculates in one go the values for all the bars, then takes the next line of script and does the same.

What i was expecting MS to do was calculate all the lines of the script for one bar, then advance to the next bar and calculate all the values again, and so on for every bar.

The results are different for each algorithm. In the example i gave, MS will calculate "a:=0;" for all the bars, then calculate "a:=ref(a,-1); +1" for all the bars and when ref(a,-1) is called, the value returned is always 0.

I'm not sure i was able to explain correctly.
Thank you all for the help.

Regards,
Jaime.

Similar Threads for: Metastock bug?
Thread	Thread Starter	Forum	Replies	Last Post
metastock	r_sajnani	MetaStock	15	15th August 2007 02:12 PM
eod for metastock	fantastic	MetaStock	2	29th July 2007 08:21 AM
metastock help?	shorttermer	MetaStock	10	16th July 2007 10:37 PM
metastock	butta_trader	MetaStock	1	16th June 2005 06:03 PM