Calculation using variable inside variable

unclesamcrazy · 04-20-2015, 08:11 AM

I have some variables which have stored numerical values like

Quote:

A1=10 A2=20 A3=25
B1=10 B2=15 B3=12 and so on

As well as another set of variables which stored these variables like

Quote:

var1=A1 var2=A2 var3=A3
var4=B1 var5=B2 var6=B3 and so on

I have to calculated things using second set of variable.
In the script, there is line

Code:

read number

user will enter number manually, suppose 500
I have to calculate using

Code:

expr 500 \* ${$var1} / 100

and output of this multiplied by

Code:

expr $above_output \* ${$var2} /100

and output of this multiplied by

Code:

expr $above_output \* ${$var3} /100

But not able to deal with two levels of variable. The calculation will be hard when it comes for three level and four level.

Please help.

Keith Hedger · 04-20-2015, 08:37 AM

If you are useing bash use the '!' indirection op to read a var like so:

Code:

A1=10
varptr=A1
echo ${!varptr}
10

You can only 'read' like this you can't use this method to set a variable, alternatly look at the eval command but it has security implications.

Code:

eval echo \$$varptr
10

allend · 04-20-2015, 09:21 AM

Do you really need the indirection? An alternative approach might be to use a function to do the calculation.

Code:

#!/bin/bash

myfunc() {
 c=$(echo "scale=3; $a * $b / 100" | bc -l)
 echo "a=$a b=$b c=$c"
}

a=500
for b in 10 20 25 10 15 12; do
 myfunc
 a=$c
done

As there is a division, I would use bc to retain numerical accuracy.

Shadow_7 · 04-20-2015, 09:47 AM

If it is bash, not specified. You can $() to execute which changes the order of precedence. $(echo $VAR), which should get treated as a value at the time of operation. Also bear in mind the $(( )) for number math in bash. And bash only deals in whole numbers, if you want to retain the decimal precision you'll have to use bc or some other application. There's also declare -i VAR to let bash know that variables are numbers and not strings which might help bash make better assumptions. If it is bash that you're doing all this in.

Keith Hedger · 04-20-2015, 10:20 AM

Quote:

Originally Posted by Shadow_7

...$(echo $VAR)...

Never thought of doing it that way

TenTenths · 04-21-2015, 02:52 AM

http://centos.tips/nesting-variables-in-bash/ has a couple of examples

rtmistler · 04-21-2015, 07:36 AM

The original problem here is this:

Quote:

var1=A1 var2=A2 var3=A3
var4=B1 var5=B2 var6=B3

This makes var1 be "A1" NOT "10"

So as Keith Hedger points out, you can establish pointers using the correct syntax to do this, or some of the other suggested techniques.

millgates · 04-21-2015, 09:24 AM

Quote:

Originally Posted by Keith Hedger

You can only 'read' like this you can't use this method to set a variable, alternatly look at the eval command but it has security implications.

Code:

eval echo \$$varptr
10

another possibility would be to use declare:

Code:

varptr=foo
declare $varptr=bar
echo $foo