Is it possible to do a bash brace expansion using a variable bound without using eval?

Question

Consider the following script.

#!/bin/bash

echo {00..99}
n=99
echo {00..$n}

The output of this script is:

00 01 02 03 04 05 06         


        

Answer 1

Not sure if this meets your requirements, as it doesn't use braces, but (with GNU seq at least) the following produces the desired output:

$ n=99
$ seq -f%02.0f -s' ' 00 "$n"
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

The "-f" option produces the zero-padding, and the "-d" uses spaces to separate, rather than newlines.

Answer 2

From the bash manual:

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

Given that variable expansion comes after brace expansion, and that there is no way to induce a different order of operations without using eval, I would have to conclude that no, there is no way to avoid using eval.

Answer 3

Does anyone know of a way to obtain the desired without without eval?

You can use seq command,

seq -w -s ' ' 0 $n

Test:

sat $ seq -w -s " " 0 $n
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99