Shell variable inside variable

Question

What is wrong in this substitution.

$ m_d_ver=0.2
$ m=mod
$ d=dom
$ echo ${$m_$d_ver}
-bash: ${$m_$d_ver}: bad substitution

Thanks,

"What is wrong in this syntax?" -- err, that it isn't the language's actual syntax? Did you find any document or reference that made you think the original proposal *would* be allowed? — Charles Duffy, Oct 20 '15 at 23:39

Charles Duffy · Accepted Answer · 2015-10-20T23:36:15.607

7

What you're trying to do is an indirect variable lookup. The syntax for that is ${!namevar}, where namevar is a variable that contains the name you actually want to evaluate. Thus:

mod_dom_ver=0.2
m=mod
d=dom
var=${m}_${d}_ver
echo "${!var}"

See BashFAQ #006.

edited Oct 20 '15 at 23:36

answered Oct 20 '15 at 23:33

Charles Duffy

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Thanks , What is this ! sign ? – user2230605 Oct 20 '15 at 23:35
It's what tells the shell to do an indirect lookup. – Charles Duffy Oct 20 '15 at 23:36
BTW, the curly braces here are needed because `_` is a valid character within a shell variable name, so if you wrote `var=$m_$d_ver`, the shell would be trying to expand `${m_}` and `${d_ver}`. – Charles Duffy Oct 20 '15 at 23:38

Shell variable inside variable

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