Without error logs (can you provide some?) it is dificult to spot the problem, however it's most probably because @ followed by a variable name is interpolated in a double quoted string, or its equivalent, so
print "My email is foo@bar.com";
tries to print the array @bar.
To avoid this, use single quotes, or escape every @ in email addresses putting a backslash before it, like this:
print "My email is foo\@bar.com";
print 'My email is foo@bar.com';
Also remember that single quotes prevent interpolation, so if yuo have \n in your string, it won't be interpreted as a newline anymore.
-- TMTOWTDI
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