Arithmetic problem with shell script

Question

I've got some issues on scripting... if someone could help me, it would be really good !

My script has:

VISITS=$((WR + RD));
SERVICE_DEMAND=$((VISITS*SERVICE_DEMAND));

And I'm getting this error:

./calc_serv_demand.sh: line 12: 0.0895406: syntax error: invalid arithmetic operator (error token is ".0895406")

Can someone help me?

I think it's because the bash works only with integer... I need to use float values, though.

thanks in advance

---

Problem solved:

VISITS=$(echo $WR + $RD | bc); echo $VISITS

SERVICE_DEMAND=$(echo $VISITS '*' $SERVICE_TIME | bc); echo $SERVICE_DEMAND

Answer 1

You can use bc to do your floating point calculations, ie

echo $WR + $RD | bc

and so on.

Answer 2

Instead of using bc , consider switching to a better programming language. Bash is simply unsuited for mathematics.

Answer 3

使用bc 在Bash中进行浮点计算 。

Answer 4

To set the precision (number of digits of the answer to the right of the decimal point), write:

WR=5
RD=7
VISITS=$[WR+RD]
SERVICE_DEMAND=.0895406
SERVICE_DEMAND=`echo "scale=5; $VISITS * $SERVICE_DEMAND" |bc -l`
echo Service Demand = $SERVICE_DEMAND

This outputs:

Service Demand = 1.0744872

The scale=5 sets 5 digits of precision; the backquotes cause the contained expression to be evaluated and the ouput (from the bc -l ) to be assigned to your variable.

Answer 5

You'll have to use an external program like bc to do floating-point math in your scripts.

Something like:

echo ($WR+$RD)*$SERVICE_DEMAND | bc