I'm currently modifying some code and wanted to add an additional argument to the code on the command line. The argument I added is the last one on the list. 
#_ Command line options & setup filenames
for ($i=0; $i<=$#ARGV; $i++)
{
    if($ARGV[$i] eq "-a")
    {
    $alignment_file = $ARGV[$i+1];
    }
    if($ARGV[$i] eq "-s")
    {
    $scorecons_file = $ARGV[$i+1];
    }
    if($ARGV[$i] eq "-i")
    {
    $image_file = $ARGV[$i+1];
    }
    if($ARGV[$i] eq "-m")
    {
    $numbered_model = $ARGV[$i+1];
    }
    if($ARGV[$i] eq "-c")
    {
    $csa_file = $ARGV[$i+1];    
    }

print "test $csa_file\n";
    
#_ Check for minimum number of command line argument variables
    if($#ARGV < 6)
    {
    Usage();
    print "NOT ENOUGH COMMAND LINE ARGUMENTS\n";
    exit 0;
    }
    
    $fh_csa = new FileHandle($csa_file, "r") || die "Cannot open CSA f
+ile: $csa_file ($!)";

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The code worked before I added the -c argument - i'm very confused as to why it doesn't work now. It doesn't seem obvious to me. Ay suggestions much appreciated! I know the error means that the $csa_file variable isn't initialized but I fail to see how. I am of course remembering to input said argument at the prompt ;)

In reply to Use of uninitialized value in concatenation (.) or string error after adding new arguement on command line by Angharad

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