Linux Bash: how to put value of $var into "$" to make it $($var) with $var bigger than 9?


I need to $ be a function of variable $a, so $($a) would be equal $value_of_a (so there would be possible to print OPTIONS of a command under number "a").

For example: if a=5 I need that bash make a call to the $5 ($5 is the OPTION5 that was when the program was typed into the Terminal; for example in

./ qq ww ee rr tt 

the "tt" is the option №5), so I need

echo $5

but I don’t know how to make it work through a and not through typing $5 manually

I’ve tried something like this




and later on even found eval:

eval b=( '$'"$a" )

so calling $b places in $b‘s place "$-->value_of_a_here<--"

but eval gives only first digit, so if a=11 then $b prints OPTION1|1, as it would be $1 (with adding 1 as a string) and not $11.

So if there would be

./ q w e r t y u i o p g
eval b=( '$'"$a" )
echo $b

then output would be


and not the


It seems that bigger that $9 one can’t make it at all. Is it possible? (and I heard that 1 line of a command with its sending OPTIONS is limited by 32 bytes, or something about so; maybe it is here?)

Could anyone help?


You need to use braces to access positional parameters above 9.

Also, bash uses special syntax for indirection:

$ set -- q w e r t y u i o p g
$ echo $11
$ echo ${11}
$ a=11
$ echo $a
$ echo ${$a}
bash: ${$a}: bad substitution
$ echo ${!a}

The 32 byte limit you read about may be referring to the "bang path". I can’t currently find the definitive reference but the Perl documentation states:

historically some operating systems silently chopped off kernel interpretation of the #! line after 32 characters

Answered By – jhnc

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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