open() not setting file permissions correctly [dup

Question

This question already has answers here:

Why are the file permissions changed when creating a file with the open system call on Linux? (3 answers)

Closed 10 months ago.

I create a file using the code below:

#include <stdio.h>
#include <stdlib.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <errno.h>
#include <fcntl.h>
#include <unistd.h>

int main()
{
    const char* filename = "./test.out";
    int fd;
    if(-1 == (fd = open(filename, O_CREAT|O_RDWR, 0666)))
    {
        perror("Error");
        errno = 0;
    }       
    else
        puts("File opened");

    if(-1 == (close(fd)))
    {
        perror("Error");
        errno = 0;
    }
    else
        puts("File closed");

    return 0;
}

I specify the mode argument as 0666, which should grant read,write access to everyone. However, an ls -l shows

-rw-r--r-- 1 kmehta users 0 2012-01-29 16:29 test.out

As you can see, write permissions are only granted to the owner of the file. I do not know why everyone else is not granted permissions correctly. chmod a+w test.out sets the permissions correctly though.

Code compiled as gcc -Wall test.c

Specs: gcc v 4.5.0 on Opensuse 11.3 64 bit

Answer 1

The mode argument to open specifies the maximum allowed permissions. The umask setting is then applied to further restrict the permissions.

If you need to make the permissions be 0666 specifically you will need to use fchmod on the file handle after the open succeeds or use umask to set the process’ permissions mask before the open.

Answer 2

Executing this code :

#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>

int main(void)
{
        int fd;
        if((fd = open("new.file",O_CREAT,S_IRWXU | S_IRWXG | S_IRWXO)) == -1)
        {
                perror("open");
                return 1;
        }
        close(fd);
        return 0;
}

on my Linux box, where umask returns 0022, gives me a file with the following attributes :

-rwxr-xr-x 1 daniel daniel 0 Jan 29 23:46 new.file

So, as you can see, the umask masks out the write bits in my case. It looks like it's the same on your system, too.