Bash bracket expansion as a Direct Product.

[Prefect]

Hello, I have a question on using bash's bracket expansion on the two sets {$SET1} and {$(ls)}. Why doesn't the following work in bash and how else might this be done short of recursion or nested loops, in bash or another language?

> SET1=a,b,c
> echo {$SET1}{3,4};
{a,b,c}3 {a,b,c}4

Thanks in advanced,
Eric.

[drl]

Hi.

Welcome to the forums.

This script shows some of the possibilities:

#!/bin/bash -

# @(#) s1       Demonstrate bash brace expansion.

echo
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version "=o" $(_eat $0 $1)
set -o nounset

echo
echo " Results, constants:"
echo {a,b,c}{3,4}

echo
echo " Results, variable:"
SET1=a,b,c
echo {$SET1}{3,4}

echo
echo " Results, variable + eval:"
SET1=a,b,c
eval echo {$SET1}{3,4}

exit 0

Producing:

% ./s1

(Versions displayed with local utility "version")
Linux 2.6.11-x1
GNU bash 2.05b.0

 Results, constants:
a3 a4 b3 b4 c3 c4

 Results, variable:
{a,b,c}{3,4}

 Results, variable + eval:
a3 a4 b3 b4 c3 c4

The expansion does not work as you wish because of the order in which various expansions are performed:

The order of expansions is: brace expansion, tilde expansion, parame-
ter, variable and arithmetic expansion and command substitution (done
in a left-to-right fashion), word splitting, and pathname expansion.

-- excerpt from man bash

so we need to have the line scanned more than once. For that, we use eval.

There is a entire section on brace expansion in the man page; the quote is just above that section.

Best wishes ... cheers, drl

[Prefect]

That great! I was afraid it wouldn't be possible.

Thanks again for replying.
Eric.