Célestin Matte [off-list ref] 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.
The new code looks better than the old one, but the log message may be
improved.
In any case, Acked-by: Matthieu Moy [off-list ref]
--
Matthieu Moy
http://www-verimag.imag.fr/~moy/
Le 08/06/2013 20:38, Matthieu Moy a écrit :> 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.
The new code looks better than the old one, but the log message may be
improved.
Is this better?
"
In Perl, '\n' is not a newline, but instead the string composed of a
backslash followed by an "n". To match newlines, one has to use the /\n/
regexp. As the output of "rev-list --first-parent" is line-oriented,
what we want here is to match newlines, and not the "\n" string.
"
--
Célestin Matte
On Sat, Jun 08, 2013 at 08:38:56PM +0200, Matthieu Moy wrote:
Célestin Matte [off-list ref] writes:
quoted
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,
use "/$variable/o".)
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 '.', $data;
If you feed that "foo.bar.baz", it does not split it into three words;
each character is a delimiter, because the dot is compiled to a regex.
The new code looks better than the old one, but the log message may be
improved.
Agreed. I think the best explanation is something like:
Perl's split function takes a regex pattern argument. You can also
feed it an expression, which is then compiled into a regex at runtime.
It therefore works to pass your pattern via single quotes, but it is
much less obvious to a reader that the argument is meant to be a
regex, not a static string. Using the traditional slash-delimiters
makes this easier to read.
-Peff
On Sat, Jun 8, 2013 at 10:57 PM, Jeff King [off-list ref] wrote:
On Sat, Jun 08, 2013 at 08:38:56PM +0200, Matthieu Moy wrote:
quoted
Célestin Matte [off-list ref] writes:
quoted
In Perl, '\n' is not a newline, but instead a literal backslash followed by an
"n". As the output of "rev-list --first-parent" is line-oriented, what we want
here is a newline.
This is right, but the code actually worked the way it was. I'm not
sure, but my understanding is that '\n' is the string "backslash
followed by n", but interpreted as a regexp, it is a newline.
Yes, the relevant doc (from "perldoc -f split") is:
The pattern "/PATTERN/" may be replaced with an expression to specify
patterns that vary at runtime. (To do runtime compilation only once,
use "/$variable/o".)
So it is treating "\n" as an expression and compiling the regex each
time through ...
I read this as saying only that static /PATTERN/ can also be a
composed /$PATTERN/. It does not indicate how string 'PATTERN' is
treated, nor does any other part of "perldoc -f split" make special
mention of string 'PATTERN'. In fact, the only explanation I found
regarding string 'PATTERN' is in my Camel book (3rd edition, page 796)
in a parenthesized comment:
(... if you supply a string instead of a regular expression, it'll be
interpreted as a regular expression anyway.)
quoted
The new code looks better than the old one, but the log message may be
improved.
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.
Sounds good to me.