UNIX for Advanced & Expert Users

I use following code

#include <fcntl.h>
#define RWRWRW (S_IRUSR | S_IWUSR | S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH)

int main(void){
umask(0);
create("foo",RWRWRW);
umask(S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH);
create("bar",RWRWRW);
exit(0);
}

the result is follows:
$umask
002
$./a.out
$ls -l foo bar
-rw------- bar
-rw-rw-rw- foo
$umask
002

My question are follows:
1. Why bar and foo have different right setting,how to get the following result?
-rw------- bar
-rw-rw-rw- foo

2.in my code,I have umask(S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH)
why umask value is still 002 after running my code ?

thank

The "umask" function enables/disables the permissions that can be set for files on a per-process basis. The permissions given to open, create, mkdir, and other functions are effectively ANDed with the parameter given to the last invocation to umask, before the operation is performed. Hence, with a umask of 0, the 'group' and 'other' permissions are stripped from RWRWRW; whereas, with your second invocation of umask, you have allowed 'group' and 'other' to be conferred read and write privileges.

As for your second question, umask operates on a per-process basis. It will only affect the process in which umask is invoked (and potentially in subsequently spawned children of that process). It will not affect the processes's parent (i.e. the shell).

umask

I know when a umask of 0,the 'group' and 'other' permissions are stripped from 'RWRWRW',but I don't understand 'user' permission is still rw- ?

Thanks