/varcalc question

[Amadeus]

I'm figuring that /varcalc assumes that the answer of the calculation has to be an integer. Therefore, if I tried something like:

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/varcalc v5 23/10

I get an error message, right?

Would it be possible to design it so that it will "round" off and still provide an answer, or else take it to a couple decimal places? (ie, perhaps /varcalcfloat)
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I'm assuming that this is what caused the error in the macro I was writing, let me know if I'm off-base.

[L124RD]

Salutations,
i get 2...
if you do /varcalc 23/10*10 you then get 20... how can you fix this you ask? its quiet simple...

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/varcalc v0 35
/varcalc v1 $v0/10
/varcalc v2 $v0%10
/echo $v1.$v2

there you go!!

[Amadeus]

Hmmm..I assumed that my problem was as I stated in my previous post when the following line wouldn't parse:

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/varcalc v5 $v1+$v11+(($v2+$v22)/10)+(($v3+$v33)/100)+(($v4+$v44)/1000)

...but perhaps that wasn't the problem. Do I need to define v5 using /varset before doing a /varcalc?

[Plazmic]

Note: there is no () handling yet
you can get around this by

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/varcalc v5 $v1+$v11+$calc($calc($v2+$v22)/10)+$calc($calc($v3+$v33)/100)+$calc($calc($v4+$v44)/1000)

At least, I think that'll work

[Draekz]

I dont suppose anyone who failed math in grade school would still be able to program macro's huh

I dunno what that was all about but it was greek to me Oh well..im still determined to become a decent macro'er soo..i'll stick with it