The stat preceding -l _ wasn't an lstat-CodePudding

Look at two examples:

$ perl -e ' print -e "registrar" && -d _ ? "YES" : "NO" '
YES

$ perl -e ' print -e "registrar" && -l _ ? "YES" : "NO" '
The stat preceding -l _ wasn't an lstat at -e line 1.

-d and -l are both file test operators. Why second does not work? The stat preceding -l _ is same as for -d _.

CodePudding user response：

Most of the file tests use stat, which follows symlinks; -l uses lstat because it makes no sense to follow one if you want to test if it itself is a symbolic link. Hence the error.

Also, as of perl 5.10, tests can be directly chained without needing to be &&ed together. Doing this with -e and -l works since perl is smart enough to pick the appropriate stat function:

$ mkdir registrar
$ perl -E ' say -e -d "registrar" ? "YES" : "NO" '
YES
$ perl -E ' say -e -l "registrar" ? "YES" : "NO" '
NO

I actually need exists and not link. Is there short hand for that?

The shorthand notation only works for and-ing all the tests together, but you can call lstat directly to see if a file exists:

$ ln -s registrar registrar-link
$ perl -E 'say lstat "registrar" && ! -l _ ? "YES" : "NO" '
YES
$ perl -E 'say lstat "registrar-link" && ! -l _ ? "YES" : "NO" '
NO
$ perl -E 'say lstat "no-such-file" && ! -l _ ? "YES" : "NO" '
NO

And in a script as opposed to a one-liner, I'd use File::stat instead of the underscore notation.