Is there simply something wrong in my code?
I think so. This line is subtly wrong: my $execTwo = "$command &> log ";You're not just re-directing there, you're also starting the command in the background. To redirect both stderr and stdout you'd need >&, not &>. So you're getting the return value of the shell, not your program.

<waffle>I may be mis-reading your code, but I think that's what's happening.</waffle>

Do not rebuke them with harsh words ... but rather lead them gently - with URLs - so that they may learn wisdom.

In reply to Re: Different values when using $? to detect seg faults by VSarkiss
in thread Different values when using $? to detect seg faults by Gmong

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