Why does open() create my file with the wrong perm

问题:

I am trying to read some text from a file and write it to another using open(), read() and write().

This is my open() for the file-to-write-to (I want to create a new file and write into it):

fOut = open ("test-1", O_RDWR | O_CREAT | O_SYNC);

This is setting file-permissions to something I don't understand at all. This is the output of ls -l:

---------T 1 chaitanya chaitanya 0 2010-02-11 09:38 test-1

Even the read permission is locked. I tried searching for this, but could not find ANYTHING. Strangely, write() still successfully writes data to the file.

Also, if I do a 'chmod 777 test-1', things start working properly again.

Could someone please let me know where I am going wrong in my open call?

Thanks!

For your reference, I have pasted the complete program below:

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <fcntl.h>

int main () {

    char buffer[512], ch;

    int fIn, fOut, i;
    ssize_t bytes;
    FILE *fp = NULL;

    //open a file
    fIn = open ("test", O_RDONLY);
    if (fIn == -1) {
        printf("\nfailed to open file.");
        return 1;
    }

    //read from file
    bytes =  read (fIn, buffer, sizeof(buffer));
    //and close it
    close (fIn);

    printf("\nSuccessfully read %d bytes.\n", bytes);

    //Create a new file
    fOut = open ("test-1", O_RDWR | O_CREAT | O_SYNC);

    printf("\nThese are the permissions for test-1\n");
    fflush(stdout);
    system("ls -l test-1");

    //write to it and close it.
    write (fOut, buffer, bytes);
    close (fOut);


    //write is somehow locking even the read permission to the file. Change it.
    system("chmod 777 test-1");

    fp = fopen ("test-1", "r");
    if (fp == NULL) {
        printf("\nCan't open test-1");
        return 1;
    }

    while (1)
    {
        ch = fgetc(fp);
        if (ch == EOF)
            break;
        printf("\n%c", ch);
    }

    fclose (fp);

    return 0;
}

回答1:

open() takes a third argument which is the set of permissions, i.e.

open(filename, O_RDWR|O_CREAT, 0666)

0666 is an octal number, i.e. every one of the 6's corresponds to three permission bits

6 = rw

7 = rwx

It's a typical pitfall. The compiler allows you to leave the permission argument away, because when you open an existing file the permission bits don't make sense. But when you forget the argument when you create a file, you get a random set of permissions, e.g. 0000 in your case (---).

回答2:

Reading http://linux.die.net/man/2/open it seems you missed the mode parameter for open:

mode must be specified when O_CREAT is in the flags, and is ignored otherwise. The argument mode specifies the permissions to use in case a new file is created.

回答3:

This question recently helped me out, so I wanted to do my part to add a bit more depth as to what's going on. Like it was stated before, you were missing the third argument to open(). However, the permissions you see aren't random; they're coming from the stack. Look at the following code snippet:

    asm("push $0");
    asm("push $0");
    asm("push $0");
    fd = open("base", O_RDWR|O_CREAT);

Note the following result:

    ----------. 1 user user 4 Feb 26 08:21 base

Let's change the first push to 1, i.e. execute permission:

    asm("push $1;push $0;push $0");
    fd = open("base", O_RDWR|O_CREAT);

and we get:

    ---------x. 1 user user 4 Feb 26 08:25 base

Change the push to 4, i.e. read permission, and mess with the other two values:

    asm("push $4;push $5;push $6");
    fd = open("base", O_RDWR|O_CREAT);

and we get:

    -------r--. 1 user user 4 Feb 26 08:27 base

Thus we can see the third value popped off the stack (first pushed) is what really matters. Finally for fun we can try 5 and then 50, which respectively result in:

    -------r-x. 1 user user 4 Feb 26 08:27 base
    ----rw----. 1 user user 4 Feb 26 08:28 base

Hope this adds some clarity!

回答4:

Actually umask() only filters permissions and does not set them. The typical umask() value is 0002 ("don't give away write permission to the world") and if your mode value in the open( "file", O_CREAT, 0777) gave all permissions, the resulting file would have 775 as its permssions.

回答5:

Not strictly relevant to the question, but the accepted answer could use this clarifying point:

There is a relationship between rwx and its numerical representation that can be seen by treating the presence of a letter as a binary 1, and its absence as a binary 0.

e.g.

rwx  <-->  111 (binary) <-->  7 (octal)

r--  <-->  100 (binary) <-->  4 (octal)

-wx  <-->  011 (binary) <-->  3 (octal) 

As a further addendum, you may now consider the chmod command:

chmod 777 filename.extension --> rwxrwxrwx permissions

777 <--> 111 111 111 <--> rwx rwx rwx

or: chmod 654 filename.extension --> rw-r-x-r--

654 <--> 110 101 100 <--> rw- r-x r--

Hope this is informative!

回答6:

you can call umask(0); system call before using open(); system call to set your choices permissions to file correctly.

回答7:

This is kind of an old thread, but I think people should be aware of the "sys/stat.h" library. This includes a bunch of symbolic constants for setting permission bits.

For example: To open a file with Read/Write permissions enabled for the user

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

open("Your/File/Path", O_RDWR | O_CREAT, S_IWUSR | S_IRUSR);

where:

S_IWUSR // Sets the Users Write bit
S_IRUSR // Sets the Users Read bit

This library includes a bunch of others, I won't list them all here but you can read up on it all here.

Of course you can put in the octal values to set these bits, however some may argue that it is poor coding practice.