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
./hello.sh 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 $5 manually
I've tried something like this
"${$a}"
this
"${0+$a}"
and later on even found eval:
eval b=( '$'"$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 $11.
So if there would be
./hello.sh q w e r t y u i o p g
a=11
eval b=( '$'"$a" )
echo $b
then output would be
q1
and not the
g
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
q1
$ echo ${11}
g
$ a=11
$ echo $a
11
$ echo ${$a}
bash: ${$a}: bad substitution
$ echo ${!a}
g
$
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