For the benefit of everyone else, here's the description of the foo and bar files in the doc:

Look at this.  Assume you are the owner of all files and directories
listed, and that your umask (creation mask) is 022.

    directory           filename        file inode          perms
    ----------          ---------       -----------         ------
    alpha               .               100                 0755
    alpha               foo             101                 0644
    alpha               bar             102                 0444

    beta                .               200                 0555
    beta                foo             201                 0644
    beta                bar             102                 0444

Consider two directories, alpha and beta, each of which have the same
two filenames, foo and bar.  alpha/foo is really file #101, but beta/foo
is really file #201 -- different files.  However, the two bar filenames
both refer to the same file, because both #102.  It's just two links
(filenames) to the same file.

When you run the command on foo, you can write to it (0644) so perl puts a copy in foo.bak and then overwrites what was in foo.*

When you run the command on bar, you cannot write to it (0444) so it requires a different strategy. Instead it renames bar to bar.bak as the backup and then writes to a brand new file bar with what would be the changes.

Does that help at all?

* Edit: having tested this myself it is not what I see for foo. I see that foo.bak has the inode of the old foo. Are you sure of your result?

Edit 2: Both operations proceed in the same way: rename the file to .bak and then create the new file.

In reply to Re: How does rename() work on read-only files? by hippo
in thread How does rename() work on read-only files? by j41r

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