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for a in ./*.flac; do
  ffmpeg -i "$a" -qscale:a 0 "${a[@]/%flac/mp3}"

I found this script a few days ago to convert all FLAC files in the current directory into the MP3 format.

What I don't understand here is the "${a[@]/%flac/mp3}" part. I think it replaces the ending flac with mp3 for the current filename. But what eactly does the [@] part do? Is it a regular expression?

marked as duplicate by jasonwryan, kenorb, grochmal, GAD3R, don_crissti Dec 19 '16 at 21:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • @jasonwryan Not a dupe. It's not %/ here, it's /% – xhienne Dec 19 '16 at 18:09
  • @xhienne it's all parameter expansion; so it is a duplicate... – jasonwryan Dec 19 '16 at 19:35
  • @jasonwryan Thanks for pointing me in the right direction. There are two very interesting links in that other question. – monkey2k Dec 19 '16 at 20:05

As you guessed it, ${var/%flac/mp3} replaces the ending "flac" (if any) in value $var with "mp3". "${var[@]/%flac/mp3}" would do the same on each element output by ${var[@]} if var was an array.

Here, since a is not an array, you can remove [@] (I assume this is an heritage from previous attempts by the programmer).


This does a simple shell expansion. The variable a will iterate through every file matching the *.flac glob.

Since each entry will be a file (and I'm presuming no spaces in the filenames), foo.flac will be the expansion of both $a and ${a[@]}.

The construct ${var/%foo/bar} will replace foo with bar at the end of a variable var. So it's replacing the extension flac with the extension mp3 in your example to provide ffmpeg with the output filename.

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