Hi Ronald

thanx!

Thanx in advance

spikkie

$? == 256 means the child called exit(1) (256>>8 = 1). Chances are that the value (1) is not significant beyond meaning an error occurred. You'd think it would have emitted a message to STDERR identifying the error.

$ perl -le'qx{date abc}; print $?'
date: invalid date `abc'
256
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One likely cause is improper escaping of arguments. Print out $cmd, check for bad quoting, and try that same command at the prompt.

$ perl -le'qx{cat some file}; print $?'
cat: some: No such file or directory
cat: file: No such file or directory
256

$ perl -le'qx{cat "some file"}; print $?'
0
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You mentioned the child is some kind of searching utility. An error could also indicate no matches were found.

$ perl -le'qx{echo foo | grep bar}; print $?'
256

$ perl -le'qx{echo foo | grep foo}; print $?'
0
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The program's man page might identify situations where it might return an error.

You mean: Why you get different exit codes? (In both cases, the *error* is 0, but on Unix you see exit code 1 and on Windows you see exit code 0). Could it be that on Windows, a shell is running between your perl program and the application, and this is "eating" the exit code of the application? Do you get the same effect if you bypass the shell?

Another thing to try is to replace your application by one where you can control the exit code directly. For instance when I execute (on Windows 2000) the program

use strict;
use warnings;
my $r=qx(set|perl -lwe "print <STDIN>; exit 1");
print "$?\n";
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I get correct the exit code 1 (i.e. 256). What do you get in this case?
-- 
Ronald Fischer <ynnor@mm.st>

Hi Ronald, Hi Ikegami,

thanx for helping solving the problem.

Yes another shell was running in between which did not return with "exit %errorlevel%"

really appreciated your help

Kindest regards,

Spikkie