I can move 5 files to somewhere using:

$ mv my-{1..5}.jpg /path/to/dir/

How can I make copy of one file by 5 times easily

# doesn't work
$ cp my.jpg my-{1..5}.jpg

Is it possible not to use a for loop?

4 Answers 4


Try this

for f in {1..5}; do cp my.jpg my$f.jpg; done

(don't have bash here to try it myself)

  • 6
    replace for f in $f with for f in $(seq 5).
    – Dan D.
    Jan 18, 2012 at 1:36
  • 1
    Is it possible not to use a for loop?
    – kev
    Jan 18, 2012 at 1:38
  • @kev - I've always seen it done this way; also, more readabe.
    – Rook
    Jan 18, 2012 at 1:48
  • 3
    @kev, what is wrong with a for loop? This is what for loops are for...
    – soandos
    Jan 18, 2012 at 2:10
  • 3
    @DanD.: seq isn't necessary in Bash. You can use for f in {1..5} or for ((f=1; f<=5; f++)) Jan 18, 2012 at 15:57

Here is a way to do it without a for loop and without the risks of using eval:

printf '%s\n' {1..5} | xargs -I {} cp my.jpg my-{}.jpg

It's still effectively a loop.


Try tee:

tee <my.jpg >/dev/null my-{1..5}.jpg

Or parallel:

parallel cp my.jpg ::: my-{1..5}.jpg

You can do it without a loop.. using tee and {} brace expansion.

EDIT: (ammended as per Dennis Williamson's comment:

For a file named "my-.jpg"

pre="my-"; suf=".jpg"
<"$pre$suf" tee "$pre"{1..5}"$suf" >/dev/null
  • 1
    It's not necessary to use eval. <"$pre$suf" tee "$pre"{1..5}"$suf" works just fine. Jan 18, 2012 at 16:08

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