return 1 when ls -l is executed on a directory without execute permission

When ls -l is run on a directory without execute permission, fail with a error value of 1 Ok? Index: ls.c === RCS file: /cvs/src/bin/ls/ls.c,v retrieving revision 1.37 diff -u -p -r1.37 ls.c --- ls.c4 Mar 2011 21:03:19 -

ls -l behavior on directories without execute permission

Hi, When ls -l is run on a directory which has no execute permissions, ls fails but the return value is 0. bash-4.2$ ls -ld /tmp/foo/ drw-r-xr-x 3 sac wheel 512 Mar 6 18:11 /tmp/foo/ bash-4.2$ ls -l /tmp/foo/ bash-4.2$ echo $? 0 bash-4.2$ I see in the traverse function:

Re: write(2) man page

On 02/24/2013 05:34 PM, Sachidananda wrote: Ted, On Sunday 24 February 2013 10:10 AM, Ted Unangst wrote: On Sun, Feb 24, 2013 at 09:12, Sachidananda wrote: Also, for the record, it's open that seeks to the end. You can open a file with O_APPEND and seek back to the beginning, and write will n