On Sat, Jun 08, 2013 at 08:38:56PM +0200, Matthieu Moy wrote:
> Célestin Matte <celestin.ma...@ensimag.fr> writes:
> > In Perl, '\n' is not a newline, but instead a literal backslash followed by
> > an
> > "n". As the output of "rev-list --first-parent" is line-oriented, what we
> > want
> > here is a newline.
> This is right, but the code actually worked the way it was. I'm not
> sure, but my understanding is that '\n' is the string "backslash
> followed by n", but interpreted as a regexp, it is a newline.
Yes, the relevant doc (from "perldoc -f split") is:
The pattern "/PATTERN/" may be replaced with an expression to specify
patterns that vary at runtime. (To do runtime compilation only once,
So it is treating "\n" as an expression and compiling the regex each
time through (though I think modern perl may be smart enough to realize
it is a constant expression and compile the regex only once). You would
get the same behavior with this:
split $arg, $data;
if $arg contained '\n'. Of course, you _also_ get the same thing if you
use a literal newline (either "\n" or if $arg contained a literal
newline), because they function the same in a regex. In other words, it
does not matter which you use because perl's interpolation of "\n" and
the regex expansion of "\n" are identical: t hey both mean a newline.
A more subtle example that shows what is going on is this:
split '.', $data;
If you feed that "foo.bar.baz", it does not split it into three words;
each character is a delimiter, because the dot is compiled to a regex.
> The new code looks better than the old one, but the log message may be
Agreed. I think the best explanation is something like:
Perl's split function takes a regex pattern argument. You can also
feed it an expression, which is then compiled into a regex at runtime.
It therefore works to pass your pattern via single quotes, but it is
much less obvious to a reader that the argument is meant to be a
regex, not a static string. Using the traditional slash-delimiters
makes this easier to read.
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