Date: Mon, 27 Aug 2001 22:25:34 -0500 From: Dan Nelson <dnelson@emsphone.com> To: FreeBSD <freebsd@XtremeDev.com> Cc: freebsd-questions@FreeBSD.ORG Subject: Re: open() in FreeBSD? Message-ID: <20010827222534.B17383@dan.emsphone.com> In-Reply-To: <20010827210707.G43076-100000@Amber.XtremeDev.com> References: <20010827210707.G43076-100000@Amber.XtremeDev.com>
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In the last episode (Aug 27), FreeBSD said:
> Hello, I have the following test program:
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <fcntl.h>
>
> int main()
> {
> int fd = open("blah", O_WRONLY|O_CREAT);
> close(fd);
> return EXIT_SUCCESS;
> }
>
> When I compiled this and run it, ls -l blah gives me:
>
> ---------x 1 freebsd freebsd - 0 Aug 27 20:46 blah*
>
> Looking through man 2 umask, the default umask is 022. Yet the file
> created by open() gives me a mask of 001? In the shell I checked the
> umask, and it indeed is at 022. If the default umask is 022, why does
> leaving out the third argument to open() not follow chmod/umask as
> implied by the man 2 open?
>
> "open requires a third argument mode_t mode, and the file is created
> with mode mode as described in chmod(2) and modified by the process'
> umask value (see umask(2))." - man 2 open
>
> In this case it wasn't modified by the process' umask value (as it is
> set to 022 and verified). fopen() works fine. Can anyone shed some
> light on this subject? I'm sure I'm missing something simple. Or
> atleast clear up the man page on what's default and what's not.
You're missing the word "requires" in that manpage you quoted :) If
you create a file with open() you *must* pass a mode, which is masked
by the current umask setting. You didn't, so the system pulled a
random value off the stack, which when masked with your umask, happened
to come out 001. Since the default umask is 022 (filter group and
other write access), if the value on the stack happened to be 023, the
resulting mode would be 001. (all values in octal because we're
dealing with modes)
--
Dan Nelson
dnelson@emsphone.com
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