I would like to understand what this means:

... a WAV file that I am using for storing data logger information.

What is "data logger information"? is it simply a stream of 16-bit littel-endian binary integer data? If so, why put a WAV header on it? (That is, why not just have a "raw" binary file of "data logger information", stored as stream of 16-bit little-endian binary data?)

In case it makes sense for you to start with a "raw" (headerless) file containing only zeros - or some other "test pattern" - the first question is: how big should the file be (how many short ints do you want)? Let's say you want 64K: 

# create a file with 65536 bytes of zeros (i.e. 32768 short int zeros)
+:

perl -e 'print "\0" x 65536' > zeros.raw

# create a file with 32768 short int zeros followed by 32768 short int
+ ones:

perl -e 'print pack("s",0) x 32768; print pack("s",1) x 32768' > zeros
+-ones.raw

[download]
To turn a raw file into a WAV file, I'd use SoX (http://sox.sourceforge.net), but it would involve making up stuff about sample-rate, etc, which might be irrelevant to what you're actually trying to accomplish: 
sox -t raw -e signed-integer -b 16 -r 8000 -c 1 zeros.raw zeros.wav

[download]
(SoX command-line usage is pretty "unique"... but like I said, do you really need it to be a WAV file?)

As for reading binary 16-bit little-endian integer values from a raw data file (e.g. one of the "*.raw" files created above): 

#!/usr/bin/perl

use strict;
use warnings;

my $infile = shift or die "Usage: $0  raw_input_file_name\n";
my $file_size = -s $infile;
open( my $fh, $infile ) or die "$infile: $!\n";
read( $fh, my $buf, $file_size );
my @ints=unpack("s*", $buf);
print scalar @ints, " short-int elements read\n";
print "  ... which seems right\n" if ( @ints == $file_size/2 );

my %h;
$h{$ints[$_]}++ for (0..$#ints);
print "distinct int values found in $infile:\n";
print "  $_ occurs $h{$_} times\n" for (keys %h);

[download]

In reply to Re: Audio::WAV data problem by graff
in thread Audio::WAV data problem by spencoid

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