Why does the command
perl -e "print qq/a\n/ =~ /$/"
print 1?
As far as I know, Perl considers $ as the position both before \n as well as the position at the end of the whole string in multi-line mode, which is the default (no modifier is applied).
The match operator returns 1 as the true value because the pattern matched. The print outputs that value.
The $ is an anchor, which is a specific sort of zero-width assertion. It matches a condition in the pattern but consumes no text. Since you have nothing else in the pattern, the /$/ matches any target string including the empty string. It will always return true.
The $ is the end-of-line anchor, as documented in perlre. The $ allows a vestigial newline at the end, so both of these can match:
"a" =~ /a$/
"a\n" =~ /a$/
Without the /m regex modifier, the end of the line is the end of the string. But, with that modifier it can match before any newline in the string:
"a\n" =~ /a$b/m
You might get this behavior even if you don't see it attached to the particular match operator since people can set default match flags:
use re '/m'; # applies to all in lexical scope
Over-enthusiastic fans of Perl Best Practices like to make a trio of pattern changing commands the default (often not auditing every regex it affects):
use re '/msx'
There's another anchor, the end-of-string anchor \Z, that also allows a trailing newline. If you don't want to allow a newline, you can use the lowercase \z to mean the absolute end of the string. These are not affected by regex flags.