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I have a set of piped commands that generates an integer expression. A sample looks like (1 +(0x1f+0x02)) I can evaluate this expression (get the resulting value), by enclosing it in $(), so:

$ echo $((1 +(0x1f+0x02)))
34

I'm interested in getting this to work in a single command, so I tried enclosing my entire series of piped commands within $(), but it seems like bash wasn't able to evaluate the expression. This snippet probably replicates the problem I am seeing:

$ TEST="(1 +(0x1f+0x02))"

$ echo "$TEST"
(1 +(0x1f+0x02))

$ echo $($TEST)
bash: (1: command not found

Not sure what I'm missing here

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1 Answer 1

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Arithmetic expansion happens within $((...)), while $(...) is a command substitution. Since there is no command called (1, which is the first word of your TEST variable's value, you get an error from the shell when it tries to use $TEST as a command in a command substitution.

You may have wanted to do

expression='(1 +(0x1f+0x02))'

echo "$(( "$expression" ))"

With versions of bash older than release 4.4, drop the inner set of double quotes. Release 4.4 was the first release of bash that allowed for the expansion of double-quoted identifiers within an arithmetic expansion.

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  • Hmm, I'm getting errors with that though, bash: "(1 +(0x1f+0x02))" : syntax error: operand expected (error token is ""(1 +(0x1f+0x02))" ") Commented Sep 27, 2022 at 20:34
  • @rookie I can't provoke that error myself. What version of the bash shell are you using on what system? Commented Sep 27, 2022 at 20:36
  • 4.3.48 on ubuntu 16 Commented Sep 27, 2022 at 20:39
  • @rookie Ubuntu 16 reached end-of-life in April 2021. You should consider upgrading. Not for the sake of this issue, but for the sake of keeping the security of your system up to date. Commented Sep 27, 2022 at 20:41
  • @rookie Also, see updated answer. Commented Sep 27, 2022 at 20:48

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