directories I
in [Shell]
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From: Stephane CHAZELAS on 9 Feb 2010 16:35 2010-02-09, 07:56(+00), Seebs: [...] > So, what do you think /usr/bin/.././bin/./../bin is? [...] Hard to tell. It depends whether /usr/bin and /usr/bin/../bin are a symlink or not, and whether that path is given to the "cd" builtin command of a POSIX shell (without the -P option) or any other command or system call. And remember that when symlinks are involved, there can be several paths to a directory (though generally only one path with only non-symlinks components). -- St�phane
From: Barry Margolin on 10 Feb 2010 00:32 In article <0.84f21d146d89e529f6a1.20100209142434GMT.87mxzisd0d.fsf(a)bsb.me.uk>, Ben Bacarisse <ben.usenet(a)bsb.me.uk> wrote: > Ed Morton <mortonspam(a)gmail.com> writes: > <snip> > > Did you notice that the answers you've got so far are for shell, not > > necessarily for perl? Since this is comp.unix.SHELL rather than > > comp.lang.PERL, by default you should expect to get shell > > answers. There are some perl users hang out here too so chances are > > you will be able to get perl-specific answers if you say up-front in > > each posting that that's what you're looking for, but there's probably > > a more appropriate NG out there where you might reach a broader > > audience of perl experts (like comp.lang.awk for awk). I don't believe > > it's comp.lang.perl as I think I remember hearing it's not getting > > used any more but I don't know what the right NG would be, maybe a > > perl user could chime in.... > > It is (or was) comp.lang.perl.misc No message in this thread was cross-posted to that group. Maybe he posted a duplicate of the message there. But his questions seemed to be about Unix in general, not specific to any particular programming language. -- Barry Margolin, barmar(a)alum.mit.edu Arlington, MA *** PLEASE post questions in newsgroups, not directly to me *** *** PLEASE don't copy me on replies, I'll read them in the group ***
From: Ed Morton on 10 Feb 2010 08:26 On 2/9/2010 11:32 PM, Barry Margolin wrote: > In article > <0.84f21d146d89e529f6a1.20100209142434GMT.87mxzisd0d.fsf(a)bsb.me.uk>, > Ben Bacarisse<ben.usenet(a)bsb.me.uk> wrote: > >> Ed Morton<mortonspam(a)gmail.com> writes: >> <snip> >>> Did you notice that the answers you've got so far are for shell, not >>> necessarily for perl? Since this is comp.unix.SHELL rather than >>> comp.lang.PERL, by default you should expect to get shell >>> answers. There are some perl users hang out here too so chances are >>> you will be able to get perl-specific answers if you say up-front in >>> each posting that that's what you're looking for, but there's probably >>> a more appropriate NG out there where you might reach a broader >>> audience of perl experts (like comp.lang.awk for awk). I don't believe >>> it's comp.lang.perl as I think I remember hearing it's not getting >>> used any more but I don't know what the right NG would be, maybe a >>> perl user could chime in.... >> >> It is (or was) comp.lang.perl.misc > > No message in this thread was cross-posted to that group. Maybe he > posted a duplicate of the message there. I think Ben was just suggesting that that would be a group specific to perl, equivalent to comp.lang.awk for awk, in response to my asking if there was such a thing. > But his questions seemed to be about Unix in general, not specific to > any particular programming language. Right, but if you read further down in his post you'd see the script he was working on was a perl script (see below). That potential confusion is why I suggested he specifically say he's looking for a perl answer if that is the case. Ed. >> How do I determine what >> the permissions are for, say, /usr/bin/ ? >> >> What are the roles of the . and .. ? >> >> The period here seems to indicate the current directory: >> >> #!/usr/bin/perl -w >> >> opendir(THISDIR, ".") or die "tja $!"; >> @allfiles = readdir THISDIR; >> closedir THISDIR; >> print "@allfiles\n"; >> >> # ./t2.pl >text1.txt
From: Phred Phungus on 10 Feb 2010 17:38 Ed Morton wrote: > On 2/9/2010 11:32 PM, Barry Margolin wrote: >> In article >> <0.84f21d146d89e529f6a1.20100209142434GMT.87mxzisd0d.fsf(a)bsb.me.uk>, >> Ben Bacarisse<ben.usenet(a)bsb.me.uk> wrote: >> >>> Ed Morton<mortonspam(a)gmail.com> writes: >>> <snip> >>>> Did you notice that the answers you've got so far are for shell, not >>>> necessarily for perl? Since this is comp.unix.SHELL rather than >>>> comp.lang.PERL, by default you should expect to get shell >>>> answers. There are some perl users hang out here too so chances are >>>> you will be able to get perl-specific answers if you say up-front in >>>> each posting that that's what you're looking for, but there's probably >>>> a more appropriate NG out there where you might reach a broader >>>> audience of perl experts (like comp.lang.awk for awk). I don't believe >>>> it's comp.lang.perl as I think I remember hearing it's not getting >>>> used any more but I don't know what the right NG would be, maybe a >>>> perl user could chime in.... >>> >>> It is (or was) comp.lang.perl.misc >> >> No message in this thread was cross-posted to that group. Maybe he >> posted a duplicate of the message there. > > I think Ben was just suggesting that that would be a group specific to > perl, equivalent to comp.lang.awk for awk, in response to my asking if > there was such a thing. > >> But his questions seemed to be about Unix in general, not specific to >> any particular programming language. > > Right, but if you read further down in his post you'd see the script he > was working on was a perl script (see below). That potential confusion > is why I suggested he specifically say he's looking for a perl answer if > that is the case. > > Ed. Thanks all for responses. I'm working up roughly the same material in several different ways. When I posted that perl script to comp.lang.perl.misc, they prettymuch all agreed that my trouble wasn't with the programming language but the unix platform. I have huge misunderstandings as a relatively recent windows refugee. So we have: dan(a)dan-desktop:~$ ls -ld /usr/bin drwxr-xr-x 2 root root 36864 2010-02-08 19:24 /usr/bin dan(a)dan-desktop:~$ My next question is how do I determine what the max path is for this directory. My gas heater passed inspection today. I feel like I'm 2 meters tall and bulletproof. -- fred
From: Phred Phungus on 10 Feb 2010 17:45
Janis Papanagnou wrote: > Phred Phungus wrote: >> Hello newsgroup, I've got a grab bag of questions that I think are >> topical here. >> >> Apparently, unix directories have permissions. How do I determine >> what the permissions are for, say, /usr/bin/ ? > > ls -ld /usr/bin > > Permissions are coded in the first column. For their meaning read > > man chmod This is what I've been banging my head against for the last few days. The clp.misc people recommend that I think of it as chmod 700 myfile My question is: how do I go the other way? I know the answer is going to involve some arithmetic: A numeric mode is from one to four octal digits (0-7), derived by adding up the bits with values 4, 2, and 1. Omitted digits are assumed to be leading zeros. The first digit selects the set user ID (4) and set group ID (2) and restricted deletion or sticky (1) attributes. The second digit selects permissions for the user who owns the file: read (4), write (2), and execute (1); the third selects permissions for other users in the file�s group, with the same values; and the fourth for other users not in the file�s group, with the same values. So for this example, what is the equivalent scalar for the chmod command: dan(a)dan-desktop:~$ ls -ld /usr/bin drwxr-xr-x 2 root root 36864 2010-02-08 19:24 /usr/bin dan(a)dan-desktop:~$ man chmod ? > >> >> What are the roles of the . and .. ? > > The current directory and the parent of current directory, respectively. > > Janis Thanks Janis. Forgive my naivete, but doesn't this mean that a directory contains itself and its parent? -- Phred |