Re: Order of printf statements vs order of actual lines printed out

in reply to Order of printf statements vs order of actual lines printed out

It's because of the error right here:

printf "\nabove first set of dashes#:  [" . #@foo . "]\n";
# The error is -------------------here------^^
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So, what's happening? By typing #@ when you meant to type $# you initiated a comment. So "@foo . "\n"; was no longer code, but rather, a comment. That made the line in question look like this:

printf "\nabove first set of dashes#:  [" . 
printf  "first set of dashes------------\n";
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Notice how you're using the concatenation operator, "." to append the output from "printf "first set of dashes------------\n";" to the string "\nabove first set of dashes#: [". Before that concatenation can occur, that second printf has to be evaluated to obtain its return value (which is "1"). And in so doing, its side effect (in the functional sense) is taking place: It's printing the "first set of dashes....." line.

Dave

In Section Seekers of Perl Wisdom