Problems with defined variables

[#1]

This a general question because it happen sometimes and I don't know how and why it happens.

Here is the question:

Suppose you define 2 variables v1 and v2 using the input box.

There are times when you type in ($(v1)) / ($(v2)) [That's v1 divided by v2] as an expression in a chart, there is no result.

But if you type try to replace one of the variables with the actual formula, then it works.

(e.g. let's say v2 = formula2. Then if I change the expression to $(v1)/(formula2) it will work.

I am sorry if this question is not clearly stated.

Just wondering if anyone have similar problem before?

[rwunderlich]

I'm not sure on your specific example without knowing the content of your variables, but here's a general explanation that may help you understand.

The $() causes the variable to be evaluated before substitution. So

  $(v1)) / ($(v2)

means the result of v1 divided by the result of v2. By contrast

  v1 / v2

means the contents (as a string) of v1 divided by the contents of v2.  So a chart dimension X with a value of 6 and:

  v1 = 2

  v2 = X * 2

$(v1) + $(v2)  =  2 + 6 * 2  =   14

v1 + v2   =  2 + 'X' * 2  = null

v1 + $(v2)  =  2 + 6 * 2  =   14

I know it doesn't quite explain your particular example, maybe you can post some example variables.

-Rob

[#3]

I tried to reproduce your concern, but was not able to. Can you share a specific scenario in which you came across this.?

Thanks,

Angad

[#4]

Thanks everyone.

I face similar situation today. And I found that my situation here related to the comments.

For example, you have a variable v1, if you write a formula and then followed by a comment out formula like this:

(some formula)

//(some formula)

When you use the variable alone in the expressions, it will work.

But if you use the variable with other variables like ($(1))/($(v2)), it will not work.

You need to change the order of the normal formula and comment out formula in order to make it work:

//(some formula)

(some formula)

Hope this would help

PC

[rwunderlich]

That's interesting. It kind of makes sense, and kind of not. If

v1 = 4 // this is a comment

v2 = 3

$(v1) / $(v2) = 4 because the comment is included.

However.

$(v1) = 4.  The comment does not appear in final result.

-Rob