(Note: Both of these examples may require GNU or BSD find and may not work as-is with the POSIX "edition" of find).

Both of the scriptlets below must be understood as excerpts from way more complex scripts, with the gist having been compacted into a few lines.
As I get the exact same results with both variants, I'm wondering where the differences (respectively: pitfalls) of either implementation may be...

  • var. 1
while IFS= read -r f; do

  echo "reading entry: "$f""

done < <(find ~/workdocs -type f -name '*.pdf' -print)
  • var. 2
while IFS= read -rd '' f; do

  echo "reading entry: "$f""

done < <(find ~/workdocs -type f -name '*.pdf' -print0)

It turned out that the output (which this question is all about here) was entirely identical in both cases. (Tested by redirecting to output log file, then diff'ing both)
Still, I am wondering if there might be any borderline cases which may actually cause the outputs from variant #1 and #2 to differ.

This is what I'd like to know about.

  • 1
    The later handles filenames that contain a newline character.
    – jordanm
    Dec 14, 2014 at 18:21
  • 2
    The echo quoting is nonsense though: echo "reading entry: "$f"" Dec 14, 2014 at 18:35
  • 1
    I didn't mean "nonsense" as "unnecessary". I meant it as nonsense. You should really understand what you have done there. It is a good general rule to quote everything in order to minimize the risk of forgetting to quote. Dec 14, 2014 at 18:41
  • 4
    @syntaxerror There is no contradiction. It is OK to always quote. The point is that in echo "reading entry: "$f"" the relevant part $f isn't quoted at all. It is nonsense, how the quotes are used in that command; not that they are used. Dec 14, 2014 at 19:04
  • 1
    echo "reading entry: "$f"" -- the variable is NOT quoted. You're echoing the concatenation of 3 strings: "reading entry: ", the unquoted $f and then an empty string "". You want to put the variable inside quotes: echo "reading entry: $f" Dec 14, 2014 at 21:36

1 Answer 1


For the sake of comprehensibility and visibility I abuse an answer for explaining the quoting part. May the SE karma forgive me...

This is (depending on the content of f) a special case, the problem is not easy to see:

> f=foo
> set -x
> echo "reading entry: "$f""
+ echo 'reading entry: foo'
reading entry: foo

The shells debug modus shows just one string. But that is not due to "nested quoting".

A change in the variable content or a change of the command line shows the problem:

> f=foo

> set -x

> echo "reading entry: " $f ""
+ echo 'reading entry: ' foo ''
reading entry: foo

> echo "reading entry: "    $f     ""
+ echo 'reading entry: ' foo ''
reading entry: foo

> f="foo    bar"
+ echo 'reading entry: foo' bar
reading entry: foo bar

If this kind of quoting worked then the spaces would be preserved.

  • Thanks for "abusing" (?) this answer! As said in my comments, it was a long-time false interpretation of yours truly of this quoting stuff. Well, of course I knew that in JavaScript (simple) nested quoting is achieved by using single quotes for the outer and double ones for the inner term. BUT I knew as well that with echo the use of single/double quotes had a totally different meaning (i. e. interpreting $... variables and other things (y/n)). So I assumed the bash way (e. g. echo) is to use double quotes twice. However, had I had a notion of how this is interpreted...oh well. Dec 15, 2014 at 1:11
  • @syntaxerror It's "abuse" because this answer does not refer to the question... I am not familiar enough with Javascript but single and double quotes (or backslashes) can be combined in bash, too: echo "The file '${file}' is empty." That is not the same like echo 'The file "${file}" is empty.' Dec 15, 2014 at 1:27
  • Yes, that's what I was just trying to point out. In JS, the choice of quotes (single/double) does not change the meaning. But in bash, it will, with usually entirely different results in resolving variables (getting literal $var output instead of their value) etc... Dec 15, 2014 at 2:12

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