How to use Bash substitution in a variable declaration

Question

I try to declare a variable x with all chars from a..x. On the command line (bash), substitution of a..x works w/o any ticks.

$ echo {a..x}
a b c d e f g h i j k l m n o p q r s t u v w x

But assigning it to variable via x={a..x} results in {a..x} as string. Only x=$(echo {a..x}) works.

The question is: Is this the proper way of assignment or do I have to do other things?

The main aim is to assign the sequence to an array, e.g.,

disks=( $(echo {a..x}) )

brace expansion fails on assignment because it produces a list and the variable only accepts a scalar value; bash is "smart" in this sense because it knows it can't assign a list to the variable so it doesn't expand it all, that's the same reason you can assign variables containing spaces without quoting them, since it's a scalar value bash doesn't apply word-splitting on the contents even when unquoted. the array however do accepts a list and in that case both brace expansion and word-splitting happens. — Samus_, Jan 02 '12 at 21:33

score 6 · Accepted Answer · answered Jan 02 '12 at 12:05

6

You can also use set (but be sure to save positional parameters if you still need them):

set {a..x}
x="$@"

For arrays, brace expansion works directly:

disks=( {a..x} )

answered Jan 02 '12 at 12:05

choroba

`d=( {a..x} ) ; echo $d` on cmdline as well as in a script results in `a` only – tuergeist Jan 02 '12 at 12:51
1

@tuergeist, try `echo ${d[@]}` or `echo ${d[*]}` http://tldp.org/LDP/abs/html/arrays.html – Aleksandra Zalcman Jan 02 '12 at 13:06
@akk: Thanks! Of course, just `d` points to the first element only. – tuergeist Jan 02 '12 at 13:09
@DominikDitoIvosevic: See `man bash`: > Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0. – choroba Apr 02 '16 at 19:51

1 Answers1