BASH:Why is printf to a var removing spaces?

Question

Why is the output of these two different? I want the spaces to be preserved.

#!/bin/bash

x="foo"
printf "%-12s *" $x

echo " "

y=$(printf "%-12s" $x)
echo $y "*"

Running it gives

foo          *
foo *

I want the second line to look like the first.

`printf` does preserve the spaces -- its the `echo` that's removing them (when unquoted). Try `echo "$y *"` instead. — costaparas, Dec 29 '20 at 00:39
Very closely related to [Capturing multiple line output into a Bash variable](https://stackoverflow.com/a/613580/15168) — the basic answer is "quote variables used in argument lists" (almost always — absolutely always if the spacing matters). — Jonathan Leffler, Dec 29 '20 at 01:01

score 2 · Answer 1 · answered Dec 29 '20 at 00:40

2

You'll need to quote the $y:

echo "${y} *"

#!/bin/bash

x="foo"
printf "%-12s *" $x

echo " "

y=$(printf "%-12s" $x)
echo "$y" "*"

➜ ./test.sh
foo          *
foo          *
➜

answered Dec 29 '20 at 00:40

0stone0

score 0 · Answer 2 · answered Dec 29 '20 at 00:57

0

Ypu can use printf too

printf '%s *\n' "$y"

answered Dec 29 '20 at 00:57

2

You can use `printf` again, but the key point is quoting `"$y"`. – Jonathan Leffler Dec 29 '20 at 01:02

2 Answers2