G'day WingedKnight,

Welcome to the Monastery.

When in doubt, it's often a good idea to find out how Perl is seeing your code with B::Deparse.

$ perl -MO=Deparse,-p -e 'print "foo, " . (1 ? "yes" : "no") . " bar";
+'
print('foo, yes bar');
-e syntax OK
[download]

$ perl -MO=Deparse,-p -e 'print (1 ? "yes" : "no") . " bar";'
(print('yes') . ' bar');
-e syntax OK
[download]

"Why isn't " bar" getting appended to the output in the second case?"

Because it's getting appended to the return value from print.

$ perl -e '$x = print "foo, " . (1 ? "yes" : "no") . " bar"; print "\n
+$x\n"'
foo, yes bar
1
[download]

$ perl -e '$x = print (1 ? "yes" : "no") . " bar"; print "\n$x\n"'
yes
1 bar
[download]

Of course, you should start all your code with:

use strict;
use warnings;
[download]

In this case, you would have received warnings:

$ perl -e 'use strict; use warnings; print (1 ? "yes" : "no") . " bar"
+;'
print (...) interpreted as function at -e line 1.
Useless use of concatenation (.) or string in void context at -e line 
+1.
yes
[download]

See "perlintro: Safety net".