Zero padding numbers in a Bash loop

Question

I'm trying to make a list with a simple bash looping

I want this:

000000 000001 000002

They give me this:

0 1 2

My shell code:

countBEG="000000" countEND="999999" while [ $countBEG != $countEND ] do echo "$countBEG" countBEG=$[$countBEG +1] done

The problem is that when you use arithmetic operations, leading zeroes are meaningless and thus stripped. You might want to work on numbers from the beginning, then pad them when you display them (with printf for example) — Aaron
– Aaron, Commented Oct 17, 2017 at 14:44
Numbers aren't zero-padded; string representations of them are, and $((...)) (don't use $[...], it's long been deprecated) simply returns the simplest representation it can for its expression. — chepner
– chepner, Commented Oct 17, 2017 at 15:36
@Aaron, pedantically, leading zeroes are highly significant for arithmetic operations: they denote octal numbers: x="08"; ((x++)) — glenn jackman
– glenn jackman, Commented Oct 17, 2017 at 16:41

123 · Accepted Answer · 2017-10-17 14:48:38Z

3

Change your echo to use printf, where you can specify format for left padding.

printf "%06d\n" "$countBEG"

This sets 6 as fixed length of the output, using zeros to fill empty spaces.

edited Oct 17, 2017 at 14:48

123

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answered Oct 17, 2017 at 14:46

Vasan

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2 Comments

Vasan Over a year ago

Thanks @123 for fixing the broken formatting. Was trying to figure out how to escape backticks :)

123 Over a year ago

4 space at the start of the line for a code-block, all other formatting is then ignored :)

Aaron · Accepted Answer · 2017-10-17 14:54:03Z

2

The following command will produce the desired output (no need for the loop) :

printf '%06d\n' {1..999999}

Explanation :

{1..999999} is expanded by bash to the sequence of 1 to 999999
the format string '%06d\n' tells printf to display the number it is given as argument padded to 6 digits and followed by a linefeed
printf repeats this output if it is given more arguments than is defined in its format specification

edited Oct 17, 2017 at 14:54

answered Oct 17, 2017 at 14:51

Aaron

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9 Comments

123 Over a year ago

why would you use g ?

Vasan Over a year ago

Just a small change, OP needs 6 digits I think.

123 Over a year ago

g will change larger number to scientific notation, not the zero padded string op wants

Aaron Over a year ago

@123 thanks for the explanation, I've changed it to d

chepner Over a year ago

Starting in bash 4.0, brace expansion will preserve leading zeros, eliminating the need for printf: {001..100} produces 001 002 003 ... 100.

|

glenn jackman · Accepted Answer · 2017-10-17 16:40:43Z

2

You're looking for:

seq -w "$countBEG" "$countEND"

The -w option does the padding.

answered Oct 17, 2017 at 16:40

glenn jackman

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4 Comments

Patrick Reis de Souza Over a year ago

That's it. Thank you brow.

anubhava Over a year ago

May I request you to vote to undelete and reopen this question please

glenn jackman Over a year ago

@anubhava, done. How many undelete votes are required to re-open?

anubhava Over a year ago

Thanks so much. It will need 1 more undelete vote. Please also vote to reopen.

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