bash: printf ("%.s ") doesn't keep whitespaces after the s

Question

I have a piece of code like this:

printf "%.s  "  $(seq 1 $count)

It actually belongs to an else condition and its job is to print out whitespaces $count times..

It works fine if I enter a string like this:

printf "%.shelloworld  "  $(seq 1 $count)

but not when i just put in whitespaces.

Any work arounds?

Your code prints whitespaces for me. Are you sure it doesn't work for you? — Sergii Dymchenko, Feb 18 '14 at 23:52
possible duplicate of [Print a character repeatedly in bash](http://stackoverflow.com/questions/5799303/print-a-character-repeatedly-in-bash) — Sergii Dymchenko, Feb 18 '14 at 23:57

score 1 · Answer 1 · answered Feb 19 '14 at 00:53

1

Works for me:

# printf "%.s  "  $(seq 1 10) | hexdump -C
00000000  20 20 20 20 20 20 20 20  20 20 20 20 20 20 20 20  |                |
00000010  20 20 20 20                                       |    |

Is this the exact command that's being entered, or is there other variable substitution happening? Because the exact thing you seem to be experiencing would happen if you didn't quote a particular variable expansion:

# frm="%.s  "; printf $frm $(seq 1 10) | hexdump -C
[no output]

Whereas:

# frm="%.s  "; printf "$frm" $(seq 1 10) | hexdump -C
00000000  20 20 20 20 20 20 20 20  20 20 20 20 20 20 20 20  |                |
00000010  20 20 20 20                                       |    |

answered Feb 19 '14 at 00:53

Sammitch

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At the moment, I have: if [ bla bla ];then printf bla elif [bla];then printf "|--" – COOLBEANS Feb 19 '14 at 01:16
@KevinMcInerney: Do you have a line that, when pasted into a fresh terminal, shows the problem? – that other guy Feb 19 '14 at 04:58
@KevinMcInerney you can run your script via `bash -x` and it will show you exactly how the shell is interpreting your script as it runs. – Sammitch Feb 19 '14 at 17:29

bash: printf ("%.s ") doesn't keep whitespaces after the s

1 Answers1