Different values for CCI if source if a variable?

[windswepttom]

Hi,
I get a different value for the CCI if I hard code the source or use a variable. How come?

function preMain() {
setPriceStudy(false);
setStudyTitle("CCI");
setCursorLabelName("CCI",0);
setCursorLabelName("CCI Var",1);
setDefaultBarFgColor(Color.blue,0);
setDefaultBarThickness(2,0);
}

var cbCCI_Var = 0;
var cbCCI = 0;
var nCCIPeriod = 14;
var nCCISource = "hlc3()";

function main() {

cbCCI = cci(14,hlc3()).toFixed(2)*1;
cbCCI_Var = cci(14,nCCISource).toFixed(2)*1;

var retArray = new Array(2);

setBarFgColor(Color.blue, 0);
setBarFgColor(Color.red, 1);
retArray[0] = cbCCI;
retArray[1] = cbCCI_Var;

return retArray;
}

Attached should be a screen shot.

How do you get a screen shot into the text?

Tom

[ACM]

Tom
That is happening because the syntax used in the cci() function in the following line of code
cbCCI_Var = cci(14,nCCISource).toFixed(2)*1;
is incorrect for what you are trying to accomplish. The cci() function expects a series object as the source parameter and if it does not find one it reverts to its default source which is close().
In your case nCCISource is a string ie "hlc3()" and not a series object so you need to evaluate it to be able to use it as the source parameter. Try replacing that line of code with the following
cbCCI_Var = cci(14,eval(nCCISource)).toFixed(2)*1;
and you should see the plot match the one returned by the other cci() function.
Alex

[windswepttom]

Thank You Alexis

I did not know about the method eval(). I have a good use for it in another application.