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Postings from December 2001
En réponse à Simon Cozens <fwp@simon-cozens.org>:
> but I guess this would be a pretty useful idiom:
>
> perl -aln0 -F"\n" -e ...
>
> (Stick an entire file into @F, a line per element.)
>
> However, if you write it as a program, instead of
>
> #!/usr/bin/perl -aln0 -F"\n"
> ...
>
> you can say
> #!/usr/bin/perl -aln0 -F
> ...
>
> because the character after -F happens to be a newline. Handy, that, eh?
I have been trying this all morning, but it doesn't seem to work on the
Windows 95 box I use at work (I didn't have a choice)...
#!/usr/bin/perl -aln0 -F"\n"
print join "<-\n->", @F;
spits out (whatever delimiter I might try, ", ' or /):
syntax error at wc.pl line 2, near "print"
Execution of wc.pl aborted due to compilation errors.
With -MO=Deparse, I don't get much:
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
}
-MO=Deparsing this (with a space added)
#!/usr/bin/perl -aln0 -F "\n"
print join "<-\n->", @F;
gives:
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
@F = split(/ "\n"\r\n/, $_, 0); # not what I wanted...
print join("<-\n->", @F);
}
wc.pl syntax OK
So I tried your no-arg trick:
#!/usr/bin/perl -aln0 -F
print join "<-\n->", @F;
and got:
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
@F = split(/\r\n/, $_, 0); # not quite what we want
print join("<-\n->", @F);
}
But when I try this, it only works with files that I explicitely wrote with
$\ = "\r\n"... Otherwise it gives a @F array holding the whole file as its
only element. Is this just me or my version of Perl?
This is perl, v5.6.1 built for MSWin32-x86-multi-thread
(with 1 registered patch, see perl -V for more detail)
Copyright 1987-2001, Larry Wall
Binary build 628 provided by ActiveState Tool Corp. http://www.ActiveState.com
Built 15:41:05 Jul 4 2001
Anyway, I'll try those under Unix, and send my (not so good) entries to Santa
tonight.
--
Philippe BRUHAT - BooK
When you run from your problem, you make it that much harder for good
fortune to catch you, as well. (Moral from Groo The Wanderer #14 (Epic))
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