UNIX for Dummies Questions & Answers

Hello All,

I was doing some work with umask and want to make sure I'm right.
I want newly created files to be executable for the group and the world, writable for the user only, and readable by the user, group, and world.

I came up with : umask 122 by assigning:
a 6 to user a 5 to group and 5 to world then subtracting 655 from 777 and getting 122?

Am i on the right track?
Thanks for looking.

Bob

umask will never allow you to create executable files. The most permissions it will allow are 666.

Not quite. The creat(const char *path, mode_t mode), open(const char *path, int flag, mode_t mode), and openat(int fd, const char *path, int flag, mode_t mode) system calls (when being used to create regular files) take the mode argument supplied by the programmer and mask off any permission bits specified by the umask. When an application (such as a shell output redirection or the vi :w command) creates a text file, it will usually have the low order bits of mode set to 0666 (RW by owner, RW by group, and RW by world). In this case if your umask is 122, the mode on the new file will be 0644 (RW by owner, R by group, and R by world). When an application (such as c99 or gcc) creates an executable file, it will usually have the low order bits of mode set to 0777 (RWX by owner, RWX, by group, and RWX by world). In this case if your umask is still 122, the mode on the new file will be 655 (RW but not X by owner, RX by group, and RX by world). Thus having a default umask of 122 is highly unusual. A much more common umask is 022 (block W access for group and world) or 027 (block W access for group and block all access for world).

Note that the umask also applies to mknod() when creating any file type and to mkdir() when creating a directory (but on directories, the X bit specifies being able to search the directory instead of being able to execute it (as it is on regular files) and write permission is used to control who is allowed to create and delete files in that directory.