8

I need to indirectly reference a variable in the bash shell.

I basically want to what you can do in make by writing $($(var)).

I have tried using ${$var} which would be the most straight forward solution in bash but then I get this error:

bash: ${$var}: bad substitution

Is there a way to do this?

What I am trying to do is to iterate over all the arguments ($1, $2, $3, ...) to a program using an iteration variable and I cannot do this without indirection.

13

If you have var1=foo and foo=bar, you can get bar by saying ${!var1}. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
    # something
done
  • Is there a way to add multiple levels of indirection? – wefwefa3 Sep 17 '15 at 12:44
  • 1
    Not that I can quickly find. The obvious guess of ${!!var1} doesn't work. However, you can always do it manually, e.g. tmp=${!var1}; echo ${!tmp}. – Tom Hunt Sep 17 '15 at 15:05
2

Using /bin/bash:

foo=bar
test=foo
echo ${!test}

# prints -> bar

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