A $ does mean end-of-string in Javascript, POSIX, Rust (if using its usual package), and Go.
I'm working with the OpenSSF best practices working group to create some guidance on this stuff. It's a very common misconception. Stay tuned.
If anyone knows of vulnerabilities caused by thus, let me know.
I wrote a compiler from Java regex to JavaScript RegExp, in which you'll find that particular compilation scheme [1].
Edit: also quoting from [2]:
> By default, the regular expressions ^ and $ ignore line terminators and only match at the beginning and the end, respectively, of the entire input sequence. If MULTILINE mode is activated then ^ matches at the beginning of input and after any line terminator except at the end of input. When in MULTILINE mode $ matches just before a line terminator or the end of the input sequence.
[1] https://github.com/scala-js/scala-js/blob/eb160f1ef113794999...
[2] https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pa...
> If MULTILINE mode is not activated, the regular expression ^ ignores line terminators and only matches at the beginning of the entire input sequence. The regular expression $ matches at the end of the entire input sequence, but also matches just before the last line terminator if this is not followed by any other input character. Other line terminators are ignored, including the last one if it is followed by other input characters.
Looks like I have some code to fix.
[1] https://docs.oracle.com/en%2Fjava%2Fjavase%2F21%2Fdocs%2Fapi...
$ php -v
PHP 8.2.7 (cli) (built: Jun 9 2023 19:37:27) (NTS)
$ php -r 'var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello world"));'
int(1)
$ php -r 'var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\nworld"));'
int(0)
$ php -r 'var_dump("hello\nworld");'
string(11) "hello
world"
...
$ php -v
PHP 7.2.26-1+0~20191218.33+debian8~1.gbpb5a34b (cli) (built: Dec 18 2019 16:09:52) ( NTS )
$ php -r 'var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello world"));'
int(1)
$ php -r 'var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\nworld"));'
int(0)
$ php -r 'var_dump("hello\nworld");'
string(11) "hello
world"
edit: https://3v4l.org/N4o8D suggests that the behavior here is identical for all versions of PHP from 4.3 to 8.3.6.
This is easily demonstrated with an example.
<?php
var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\n"));
int(1)
<?php
var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\nworld"));
int(0)
<?php
var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\n\n"));
int(0)
In PHP, the following is considered true:
> var_dump(preg_match("/^[a-z0-9 ]+\$/", "hello\n"));
That is clear proof that "$" does NOT just match the end of the string; it also accepts an extra newline at the end of the string. In PHP you need to use \z if you want to match the end of the string, or use the "D" flag when using "$".
That definition of "$" is often reasonable when you read files a line-at-a-time from a file, which is why Perl changed its definition. However, PHP is often used for server-side web applications. In this case, you are often NOT reading a line-at-a-time from a file. In such cases, allowing an extra newline at the end could be disastrous. The MediaWiki code (written in PHP) deals with this by adding the "D" flag when it uses "$", but I'm not sure it always uses it, and I doubt all PHP programs use this flag when they should.
I don't think Java should be in your first list, though? Pattern.matches("^foo$", "foo\n") returns false.
If that's true, then I fear the answer for Java may vary. The O'Reilly book on Regular Expressions, and the JDK documentation for version 21, say clearly that $ permits an optional \n at the end. The Java 8 documentation is murky, and maybe Java 8 is different.