How to use output of FFMPEG as command line argument to an executable

Question

I have a .wmv file which I want to convert to .wav file and I am using ffmpeg for the same, the command is as follows:

ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav pipe:1

the pipe:1 outputs the output wave file in STDOUT. I want to capture that wave file from STDOUT and pass it as a command line argument to my executable called foo. I want to do the conversion from wmv to wav on the fly rather than saving the .wav file. Things I have tried are as follows but none of them seem to work:

./foo $(ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav pipe:1)

./foo $(<(ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav pipe:1))

ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav pipe:1 | xargs ./foo

ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav - | ./foo -

What about using bash process redirection? Your second example, but with just <(ffmpeg ...), and not the $( and ) ? The result of that would look something like ./foo /dev/fd/63, where the fd was populated by the output of the command. — ghoti
– ghoti, Commented Feb 22, 2017 at 21:59
ffmpeg -i file.wmv -ar 8000 -sample_fmt s16 -f wav - | ./foo /dev/stdin is another option, subject to most of the caveats relevant to process substitution (FIFO, non-seekable, &c) — Charles Duffy
– Charles Duffy, Commented Feb 22, 2017 at 22:06

chepner · Accepted Answer · 2017-02-22 21:55:08Z

2

I think your second attempt is close:

./foo <(ffmpeg ...)

The process substitution expands to a name, which foo can open like any other file. The contents of this "file" are the output from ffmpeg.

Note, however, this will not work if foo expects a seekable file; the output of ffmpeg is still a stream which is not buffered in memory. If foo expects to be able to move back and forth through the file, you need to use an actual file. (ffmpeg ... > tmp; ./foo tmp)

In the first case, you are trying to pass the contents of the output stream as an argument, similar to ./foo "$(cat some.mp4)".

In the third case, you are treating the output of ffmpeg as the name of a file.

The fourth case would work if foo were written to understand - as a stand-in name for standard input. That is a program convention, not something provided by the shell.

edited Feb 22, 2017 at 21:55

answered Feb 22, 2017 at 21:36

chepner

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2 Comments

pku Over a year ago

Thanks. but I am little confused as to how the <(ffmpeg ...) syntax serves as a command line argument as opposed to stdin.

chepner Over a year ago

Process substitution is, in some sense, a way of creating a temporary file. Try echo <(:) (: being the do-nothing command). The output is the name of the "file" that the process substitution creates.

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