Sponsored Content
Full Discussion: umask conundrum
Top Forums UNIX for Dummies Questions & Answers umask conundrum Post 302670025 by Scott on Wednesday 11th of July 2012 01:52:24 PM
Old 07-11-2012
No. You can't use umask to create executable files.

I'm not sure what the second part of your question is asking "Now it starts at 777 for dirs and 666 for files..."
 

7 More Discussions You Might Find Interesting

1. UNIX Desktop Questions & Answers

Unix Grep Conundrum - Not for Noobies

Help, I have been stuck on this issue for weeks. I am a unix noobie. I have a very long string and within that string I am trying to get proc file names ie PROCNAME1=SOME_FILENAME_UPDTBASE.SQL There is a space on either side. I can't for the life of me peel out the proc name:... (13 Replies)
Discussion started by: owenian
13 Replies

2. Shell Programming and Scripting

Sed pattern space/looping conundrum

Although my sed skills are gradually developing, thanks in large part to this forum, I'm having a hard time dealing with pattern space and looping, which I suspect is what I'll need a better handle on to figure out my current issue, which is converting a multi line file like this: ... (4 Replies)
Discussion started by: tiggyboo
4 Replies

3. Shell Programming and Scripting

Conundrum - Flexible way to strip extension

Hi, First post here. I have something that may prove to be difficult. I have the following files: Example1.0.0.tar.gz Example2.tar Example3.zip Example4.0.0.0.0.0.bzip2 I need to remove the file extensions and store as a variable so they look like this: Example1.0.0 Example2... (3 Replies)
Discussion started by: Spadez
3 Replies

4. Shell Programming and Scripting

sudo scripts conundrum

hello; Got a problem running monitoring scripts using sudo ssh.. Mgmt decided to take away root sudoers access.. so most of the scripts ran as: sudo ssh $BOX ... Now I need to run them as: echo $my_pw | sudo -S -l my_user_id $BOX ... I tried this but not working.. Any wisdom/tricks... (3 Replies)
Discussion started by: delphys
3 Replies

5. Red Hat

Physical Volume Create Conundrum

I want to start by saying I already resolved my issue but I want to understand why I am seeing what I am seeing. I have a server with a RAID controller two 500GB drives and six 600GB drives. The two 500GB drives are mirrored and have the OS installed on them. The six 600GB they wanted set as... (4 Replies)
Discussion started by: scotbuff
4 Replies

6. IP Networking

iptables conundrum

Ok, if youre reading this prepare yourself.(debian based os) so im trying to do this routing with ip tables, i need to forward/SNAT traffic from 192.168.111.1 to 10.10.10.250, the 192.x.x.x ips are being shoved into a honeyd like program called inetsim so its offline, 10.10.10.125 is connected... (3 Replies)
Discussion started by: Shocco
3 Replies

7. OS X (Apple)

Help in explaining this echo conundrum.

OSX 10.12.3, default bash terminal. Consider this code and note it is calling 'sh' inside the code... #!/bin/sh echo '1\n2\n2\n3\n5' > /tmp/text hexdump -C /tmp/text /bin/echo '1\n2\n3\n4\n5' > /tmp/text hexdump -C /tmp/text Now view the interactive mode below, note the underlying shell is... (6 Replies)
Discussion started by: wisecracker
6 Replies
UMASK(2)						      BSD System Calls Manual							  UMASK(2)

NAME
umask -- set file creation mode mask SYNOPSIS
#include <sys/types.h> #include <sys/stat.h> mode_t umask(mode_t numask); DESCRIPTION
The umask() routine sets the process's file mode creation mask to numask and returns the previous value of the mask. The 9 low-order access permission bits of numask are used by system calls, including open(2), mkdir(2), mkfifo(2) and mknod(2) to turn off corresponding bits requested in file mode. (See chmod(2)). This clearing allows each user to restrict the default access to his files. The default mask value is S_IWGRP|S_IWOTH (022, write access for the owner only). Child processes inherit the mask of the calling process. RETURN VALUES
The previous value of the file mode mask is returned by the call. ERRORS
The umask() function is always successful. SEE ALSO
chmod(2), mkdir(2), mkfifo(2), mknod(2), open(2) STANDARDS
The umask() function call is expected to conform to IEEE Std 1003.1-1988 (``POSIX.1''). 4th Berkeley Distribution June 4, 1993 4th Berkeley Distribution
All times are GMT -4. The time now is 02:04 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy