[SOLVED] set a variable to a combination of two other variables

nesrin · 07-27-2011, 02:07 PM

Hello,

I want to combine two variables and create a third one in bash script. But having a problem. the code looks like:

Quote:

a=0
b=0
c=$a"_"$b

echo $c
#the result is 0_0

let a="$a+1"
echo $a
#the result is 1

echo $c
#the result is 0_0 but I want to have 0_1

the problem is that variable c is not changing although variables a and b are changed. Any idea what would be the correct syntax to get that?

I would be grateful for some help.
thanks.

crts · 07-27-2011, 02:19 PM

Hi,

I think you want something like this:

Code:

a=0
b=0
c='${a}_${b}'
eval echo "${c}"
0_0
a=1
eval echo "${c}"
1_0

You have to use 'eval' to get the desired results.

Diantre · 07-27-2011, 02:21 PM

When you assign variable c, the shell expands the values of a and b into c. If the value of a is changed later on the script, the variable c will have the previous value of a.

You'll have to perform the arithmetic on a and b first, then assign them to c:

Code:

a=0
b=0

a=$((a+1)) # a is 1
b=$((b+2)) # b is 2
c="${a}_${b}" # c is 1_2

nesrin · 07-28-2011, 09:23 AM

thank you very much for your reply.
@Diantre: I have quite a long script and a & b are updated very often, that is why I want to define c once at the beginning. But worst case I will follow that idea.
@crts:eval works well in this example, but I actually have a little more complicated script. Since I thought the problem must be about how I define the variable c (not about how I call the variable c), I did not include in previous post my 4. variable var, which is defined with awk function.

Code:

a=0
b=0
c='${a}_${b}'
let b="$b+1"

var=$(awk -v c=${c} '$3 == c{print $2}' textfile)
echo $var
#in this case I dont get the correct entry on the textfile

Is it possible to get an updated version of variable c in awk function?

catkin · 07-28-2011, 09:31 AM

c='${a}_${b}' sets c to the literal ${a}_${b}. The variables would be substituted if the were within double quotes: c="${a}_${b}"

crts · 07-28-2011, 09:37 AM

Well,

I do not know what exactly you are trying to achieve but maybe you could do it all in awk? This would eliminate such ugly constructs like:

Code:

var=$(awk -v c=$(eval echo ${c}) '$3 == c{print $2}' file) # not really recommended; there is probably a better way

Maybe you can provide a bit more insight about your goal.

crts · 07-28-2011, 09:39 AM

Quote:

Originally Posted by catkin

c='${a}_${b}' sets c to the literal ${a}_${b}. The variables would be substituted if the were within double quotes: c="${a}_${b}"

Hi catkin,

that is exactly not what the OP wants. He wants to "emulate" a pointer to $a and $b, so that $c will reflect changes of either one of them dynamically.

catkin · 07-28-2011, 09:44 AM

Quote:

Originally Posted by crts

Hi catkin,

that is exactly not what the OP wants. He wants to "emulate" a pointer to $a and $b, so that $c will reflect changes of either one of them dynamically.

Thanks for correcting my lazy reading of the thread crts

Nominal Animal · 07-28-2011, 04:45 PM

Shells do not support deferred evaluation, as far as I know; there is no way to define a variable that refers to other variables, with the references resolved at use time (and not at definition time).

I do know of two mechanisms you can use to emulate such variables, though. Assuming you use Bash or a POSIX shell (dash, for example):

Use a function that outputs the desired expression.
Instead of $name or ${name} you have to use $(name) (or `name`.
Code:
```
    c () {
        echo -n "${a}_${b}"
    }

    # Use $(c) or `c` instead of ${c} or $c.
```
Define a helper variable as the assignment expression. Evaluate the helper variable whenever the components may have changed, before using the assigned variable.
While you can use the variable normally, you'll need to eval "$Update" (if Update contained the helper assignment expression) before using the variable in an expression, to make sure all changes have propagated properly:
Code:
```
    Update='c="${a}_${b}"'

    # Run
    #    eval "$Update"
    # before using $c or ${c} to make sure all
    # changes in a and b are reflected in c.
```

It is often easier to modify the code that modifies the constituent variables, and add a "side effect" that updates the derivative variables. This is also surprisingly efficient, only adding a function call overhead to each assignment:

Code:

    EvalC='C="${A}_${B}"'

    SetAB () {
        A="$1"
        B="$2"
        eval "$EvalC"
        export A B C
    }

    SetA () {
        A="$*"
        eval "$EvalC"
        export A B C
    }

    SetB () {
        B="$*"
        eval "$EvalC"
        export A B C
    }

    # To set A to "Foo Bar" and B to "Alice-Bob", use
    #     SetAB "Foo Bar" "Alice-Bob"
    # or
    #     SetA "Foo Bar"
    #     SetB "Alice-Bob"
    # and C will always reflect the changes automatically.

There are obviously many variants of the above, and a number of tweaks you can do to make any of the above schemes better fit your needs.

Keith Hedger · 08-16-2011, 08:02 PM

Why not use an array ie

Code:

array=( 0 0 )
echo ${array[0]}_${array[1]}
0_0
((array[1]=array[1]+1))
echo ${array[0]}_${array[1]}
0_1