D
Dr. Peter Dintelmann
Hi,
perl -le 'print 4==7 ? 0 : 1'
produces "1" as output and
perl -e 'die 4==7 ? 0 : 1'
produces "1 at -e line 1.". But
perl -e 'exit 4==7 ? 0 : 1'
sets the exit code to 4 instead of 1 (using brackets around the
argument of exit() changes this).
Is there any reason for this behaviour?
Peter Dintelmann
perl -le 'print 4==7 ? 0 : 1'
produces "1" as output and
perl -e 'die 4==7 ? 0 : 1'
produces "1 at -e line 1.". But
perl -e 'exit 4==7 ? 0 : 1'
sets the exit code to 4 instead of 1 (using brackets around the
argument of exit() changes this).
Is there any reason for this behaviour?
Peter Dintelmann