perl - Why does a match against regex $ return 1 when the input string contains a newline? -

why command

perl -e "print qq/a\n/ =~ /$/" 

print 1?

as far know, perl considers $ position both before \n position @ end of whole string in multi-line mode, default (no modifier applied).

the match operator returns 1 true value because pattern matched. print outputs value.

the $ anchor, specific sort of zero-width assertion. matches condition in pattern consumes no text. since have nothing else in pattern, /$/ matches target string including empty string. return true.

the $ end-of-line anchor, documented in perlre. $ allows vestigial newline @ end, both of these can match:

"a"   =~ /a$/ "a\n" =~ /a$/ 

without /m regex modifier, end of line end of string. but, modifier can match before newline in string:

"a\n" =~ /a$b/m 

you might behavior if don't see attached particular match operator since people can set default match flags:

use re '/m'; # applies in lexical scope 

over-enthusiastic fans of perl best practices make trio of pattern changing commands default (often not auditing every regex affects):

use re '/msx' 

there's anchor, end-of-string anchor \z, allows trailing newline. if don't want allow newline, can use lowercase \z mean absolute end of string. these not affected regex flags.