Re: [FWP] $_=$l[rand@l]; (fwd)

[Ronald J Kimball <rjk@linguist.dartmouth.edu>]

On Thu, Jul 29, 1999 at 12:14:59PM -0400, John Porter wrote:
> Earlier I wrote:
> > 
> > Kevin Reid wrote:
> > > Something to try shortening:
> > > 
> > > @ARGV='iMax:Documents:Info:lwall-quotes.txt';$/='%%
> > > ';@l=<>;$_=$l[rand@l];chomp;print
> > 
> > Two characters:
> > 
> > @ARGV='iMax:Documents:Info:lwall-quotes.txt';$/='%%
> > ';@l=<>;print$l[rand@l]=~/(.*)/
> 
> To which Ronald J Kimball responded by saying that the
> semantics of /(.*)/ is different because $/ has been
> changed.

Actually, I meant the semantics of the whole block of code, not of the
regular expression.  Sorry about the confusion.


> I was not able to confirm that assertion; my tests
> showed /(.*)/ still considers . (dot) to exclude \n
> specifically, not $/.  Can anyone shed some light on
> this?  Is it a perl version difference?
> 

That's my point.  Unlike the original, your code only accounts for
newlines, and it only prints the first line of the selected block, without
a trailing newline.

The original code prints all the lines of the selected block, with a
trailing newline.

Try these two code blocks, where I have replaced reading from a file with
reading from DATA.


$/='%%
';@l=<DATA>;$_=$l[rand@l];chomp;print
__END__
This is the first block.  It is one line.
%%
This is the second block.  It is two lines.
Here is the second line.
%%


$/='%%
';@l=<DATA>;print$l[rand@l]=~/(.*)/
__END__
This is the first block.  It is one line.
%%
This is the second block.  It is two lines.
Here is the second line.
%%

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