[Tutor] subprocess.call warning

Steven D'Aprano steve at pearwood.info
Mon Oct 4 00:47:11 CEST 2010

On Mon, 4 Oct 2010 06:44:54 am Norman Khine wrote:
> hello, from the docs http://docs.python.org/library/subprocess.html i
> see there is a WARNING about deadlock when using the subprocess.call.
> in my code i have this
> http://pastie.org/1197024

The warning says:

    Like Popen.wait(), this will deadlock when using stdout=PIPE 
    and/or stderr=PIPE and the child process generates enough 
    output to a pipe such that it blocks waiting for the OS pipe 
    buffer to accept more data.

Since you don't set either stdout or stderr to PIPE, you shouldn't have 
to worry about a deadlock.

> the first calls the 'sox' library which joins all the .wav files into
> one file and then i use the 'wav2swf' library to create a SWF output
> of the file.
> can the code be improved?

Code can always be improved :-) What do you consider an improvement? 
Easy to maintain in the future? Faster? Smaller?

What does get_abspath do? It sounds like it *may* be your own version of 

I'm not sure why you have a list called "imgtext" that doesn't contain 
text. It seems to be a list of partial file names rather than "image 

Your code, as given, can't work, because you call isupper() on integers 
1 2 and 3. That will fail.

I'd probably prefer to generate the filenames differently:

def fix_filename(fname, base=sound_path):
    fname = str(fname)
    if fname.isupper():
        fname = "upper_%s" % fname.lower()
    return os.path.join(base, fname + '.wav')

and then call it:

filenames = ['A', 'b', 'c', 'D', 1, 2, 3]
for name in filenames:
    name = fix_filename(name)

Steven D'Aprano

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