So reading perlre, a $ matches:

• at the end of the string (your first case)
• or before newline at the end of the string (what happens in your second case, leaving the newline out of $1)
• before any newline if you've used /m

If you use (say) Regexp::Debugger it might help stepping through and watching.

Well... I would buy it "by default" or with /m qualifier, but I did add a /s qualifier. And with that I think \n is supposed to be treated just like any other character. Should it not?

s Treat string as single line. That is, change "." to match any character whatsoever, even a newline, which normally it would not match.

You're not taking into account the non-greediness. To accommodate matching the $ (which is before the newline) $1 holds 'foo'. If you also want to match the terminal newline, use \z instead of $:

$ perl -e '$_="foo\n"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo|

$ perl -e '$_="foo\n"; print "1=|$1|\n" if m/(fo.+?)\z/s'
1=|foo
|
[download]

Your comparison with the first one-liner is not comparing apples with apples:

$ perl -e '$_="foo\nbar"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo
bar|

$ perl -e '$_="foo\nbar\n"; print "1=|$1|\n" if m/(fo.+?)$/s'
1=|foo
bar|

$ perl -e '$_="foo\nbar\n"; print "1=|$1|\n" if m/(fo.+?)\z/s'
1=|foo
bar
|
[download]

I also second ++Fletch's recommendation to use Regexp::Debugger. This allows you to step through the matching process and see exactly what's happening. I often use it myself.

— Ken

From the title and body, it's possible you think greediness (or non-greediness) can affect *if* a match occurs, but that's not the case.

(In retrospect, you might already realize that. If so, just ignore this.)

the $ is not inside your $1 group, so why should \n be included in the second print?

anyway $ is a zero-length assertion

never mind, fletch already gave the perfect answer.

• "before newline at the end of the string"