I'm writing a program that writes output to a file. If this file doesn't exist, I want to create it.
Currently, I'm using the following flags when calling open:
O_WRONLY | O_CREATE
However, when this creates the file, it doesn't give me any permissions to write to it...
How can I use open so that it creates a file if it doesn't exist, but will create it with necessary permissions when needed?
Thanks!
You probably need the third argument. For example:
open('path',O_WRONLY|O_CREAT,0640);
Just use the optional third argument to open:
int open(const char* pathname, int flags, mode_t mode);
so like this:
open("blahblah", O_CREAT | O_WRONLY, S_IRUSR | S_IWUSER | S_IRGRP | S_IROTH);
See man open(2).
On Linux there's a third argument you can use to pass permissions. S_IWUSR should be the flag to give you write permissions, but in practice you'll probably want to use more flags than just that one (bitwise or'd together). Check the manpage for a list of the permission flags.
From the manual:
O_CREAT
If the file exists, this flag has no effect except as noted under O_EXCL
below. Otherwise, the file shall be
created; the user ID of the file shall
be set to the effective user ID of the
process; the group ID of the file
shall be set to the group ID of the
file's parent directory or to the
effective group ID of the process; and
the access permission bits (see
) of the file mode shall
be set to the value of the third
argument taken as type mode_t modified
as follows: a bitwise AND is performed
on the file-mode bits and the
corresponding bits in the complement
of the process' file mode creation
mask. Thus, all bits in the file mode
whose corresponding bit in the file
mode creation mask is set are cleared.
When bits other than the file
permission bits are set, the effect is
unspecified. The third argument does
not affect whether the file is open
for reading, writing, or for both.
Implementations shall provide a way to
initialize the file's group ID to the
group ID of the parent directory.
Implementations may, but need not,
provide an implementation-defined way
to initialize the file's group ID to
the effective group ID of the calling
process.
So it seems you need to pass a third argument specifying the desired file permissions.
Note that under POSIX (Unix, Linux, MacOS, etc) you can open and create a file with any permissions you choose, including 0 (no permission for anyone), and yet still write to the file if opened for writing.
Related
I am trying to implement my own shell in which i have to make a shell feature by which i will able to append text to file by using >>.here is my code....
int filedesc = open(inputargs[limit-1],O_WRONLY | O_APPEND);
//printf("%s\n",inputargs[limit-1]);
if(filedesc < 0) {
printf("Error opening file\n");
}
else{
dup2(filedesc,1);
}
filedesc always returning -1.
Transferring an edited form of my relevant comments into an answer.
Using O_APPEND is correct — and you do need O_WRONLY too — and maybe you should add O_CREAT in case it doesn't exist yet (but you definitely don't want O_TRUNC). I suppose O_RDWR would also work (in place of O_WRONLY), but it is not a good idea.
Does the file exist to be appended to? If not, add O_CREAT and a file mode argument. Tradition says "use 0666" (aka S_IRUSR | S_IWUSR | S_IRGRP | S_IWGRP | S_IROTH | S_IWOTH — there's a reason people prefer typing the octal number!). I'd be sorely tempted to not grant write to group or others (using 0644), even though you can (and I do) use umask 022 to suppress those permissions.
What do you get from perror(inputargs[limit-1]); (instead of or as well as the printf() message? That will tell you why the system thinks it can't open the file. And don't forget that the system is right — it can't open the file. Your code has a problem — it is not yet clear what that is. It might be EPERM (no permission); it might be ENOENT (no such file or directory); it might be something else, but those two are the most likely.
Getting "permission denied" from perror(inputargs[limit-1]);
So, you don't have permission to write on the file — what are you going to do about that? Change the file name to one you have permission to write on, or set the permissions on the file you have so that you can write (append) to it? Or something else? If you, somewhere along the line, create the file using O_CREAT but don't specify the third (optional) argument to open(), you get quasi-random weird permissions on the file that you create. What does ls -l file-to-append-to tell you? (Note that the third argument is mandatory when O_CREAT is one of the options used in the call to open() — it is optional and ignored otherwise.)
Note that you should always write error messages in a shell to standard error and not to standard output. Even if there's I/O redirection on standard error, the shell still writes the error messages to standard error, and the redirection points it where the user wanted it to go.
I have some trouble understanding the arguments in the open function, specifically used in the context of creating an output file. I do not quite understand the roles of flags and file permissions (the 2nd and 3rd arguments in the function). For instance, if I have the file permission 00200 (user has write permission), and the flag O_RDONLY (Read only), then can I read the file or write the file?
The signature of open is as follows:
int open(const char *pathname, int flags, mode_t mode);
There are three sets of "permissions" at play: The permissions of the file itself, the flags, and the mode.
The permissions of the file itself (e.g. 00200 meaning only user can write) specify what the operating system allows a program to do.
When you specify the flags, you indicate what you want to do with the file. For example, if the file is readonly to you (e.g. rwxr-xr-x and you're not the owner), you will be allowed to open the file with O_RDONLY. If you attempt to open the file with O_RDWR or O_WRONLY, you will receive an EPERM (operation not permitted) error in errno.
The mode parameter is only relevant when you create a new file, such as when you open a file that doesn't exist1 and the flag O_CREAT is specified. The file is created on the filesystem and its permissions are given by mode & ~umask (see man 2 umask for more details).
1 Of course, the containing directory must exist and you must have write+exec permissions on that directory.
If you look at this code block below by taking into consideration the last parameter "0", Does write line work properly ?
filename = argv[1];
string = "Example string";
if (stat(argv[1], &buf) != 0)
{
fd = open(filename, O_WRONLY | O_CREAT, 0);
if (fd < 0)
{
perror(filename);
exit(1);
}
write(fd, string, strlen(string));
close(fd);
}
else
{
print("%s file exists\n", filename);
}
From the manpage:
mode specifies the permissions to use in case a new file is created. This argument must be supplied when O_CREAT is specified in flags; if O_CREAT is not specified, then mode is ignored. The effective permissions are modified by the process's umask in the usual way: The permissions of the created file are (mode & ~umask). Note that this mode applies only to future accesses of the newly created file; the open() call that creates a read-only file may well return a read/write file descriptor.
The following symbolic constants are provided for mode:
S_IRWXU 00700 user (file owner) has read, write and execute permission
S_IRUSR 00400 user has read permission
S_IWUSR 00200 user has write permission
S_IXUSR 00100 user has execute permission
S_IRWXG 00070 group has read, write and execute permission
S_IRGRP 00040 group has read permission
S_IWGRP 00020 group has write permission
S_IXGRP 00010 group has execute permission
S_IRWXO 00007 others have read, write and execute permission
S_IROTH 00004 others have read permission
S_IWOTH 00002 others have write permission
S_IXOTH 00001 others have execute permission
So, specifying a mode of zero, you will create a file with the permissions of 0 & ~umask, i.e. a file without any permissions.
What exactly the filesystem makes of this is not in the domain of the open() or write() functions.
It is valid,
This is from open(2) Linux manual pages
The mode argument specifies the file mode bits be applied when a new file is created. This argument must be supplied when O_CREAT or O_TMPFILE is specified in flags; if neither O_CREAT nor O_TMPFILE is specified, then mode is ignored. The effective mode is modified by the process's umask in the usual way: in the absence of a default ACL, the mode of the created file is (mode & ~umask). Note that this mode applies only to future accesses of the newly created file; the open() call that creates a read-only file may well return a read/write file descriptor.
In theory then, your access to the file will be valid until you call close() as I understand the part I highlighted in the above excerpt.
Interesting question. POSIX says:
The argument following the oflag argument does not affect whether the file is open for reading, writing, or for both.
Which means that since you're handling the error return from open, if you reach the write line the behavior is well defined.
To expand a bit why this works. On most filesystems on unix-like systems, the meta-data related to a file should not affect already open file descriptors. You can for example remove a file that you have opened. This is in fact done quite commonly with temporary files, so that you don't need to remember to delete them on exit. The same applies to permissions or even ownership of the file. In fact, you can chroot while holding a file open and you can still write to it without actually being able to see it. You can even use file descriptor passing to give an open file descriptor to another process that wouldn't be allowed to open that file. This is quite commonly used for privilege separation. The permissions you had when creating a file descriptor are valid regardless of the changes to permissions later. So your question is a very interesting edge case because it asks if the filesystem permissions of the file are set before or after we create a file descriptor for it and POSIX seems to be clear on that.
I can only think of two exceptions to that right now. First is when someone forcibly remounts a filesystem to read-only in that case the kernel will go through horrifying gymnastics to invalidate your file descriptor which will make all its operations fail. Second one is AFS where your permissions are actually checked when you close the file (or, when the last user of the file on your local system closes it which sends it to the server), which leads to hilarious problems where your time-limited access tokens were valid when you opened a file but aren't valid any longer when you close it. This is also why close returns errors (but that's another rant).
This is why I mentioned error handling above. Even though POSIX says that it should not have an effect, I could see AFS or certain other file systems refusing to open such a file.
I'm creating a file in C using the following code:
int outfd = open(arg,O_CREAT|O_TRUNC|O_WRONLY, f_per);
f_per being the file permission numbers.
Setting f_per to 0644, executing the code and doing an ls -l gives me the (output) file permissions set as -rw-r--r-- which is expected. However, setting things to 0777 gives permissions as -rwxrwxr-x instead of -rwxrwxrwx. Any idea why this happens?
As per the POSIX page for the open call, under O_CREAT:
... the access permission bits of the file mode shall be set to the value of the argument following the oflag argument taken as type mode_t modified as follows: a bitwise AND is performed on the file-mode bits and the corresponding bits in the complement of the process' file mode creation mask.
The mode creation mask (or umask) can be considered a subtractive one. For example, if you attempt to create a file with permissions rwxrwxrwx/0777 when your file mode creation mask is -------w-/0002, you'll actually end up with:
rwxrwxrwx
& rwxrwxr-x (complement of -------w-)
=========
rwxrwxr-x
This appears to be the situation you are encountering.
If you want to actually create a file of the specific permissions, you can temporarily disable the umask by setting it to zero (and restoring it afterwards), something like:
mode_t oldmask = umask(0); // needs sys/stat.h
int outfd = open(arg, O_CREAT|O_TRUNC|O_WRONLY, 0777);
umask(oldmask);
I call the following code in C:
fileCreatefd = mkstemp(fileName);
I see that the file is created with permissions 600 (-rw-------). I want to create this temp file as -rw-rw-rw-
I tried playing around with umask but that only applies a mask over the file permissions -- at least thats my understanding. So how can i create a file with permissions 666?
Thanks
You cannot create it 0666 with mkstemp. You can change the permissions afterwards, if that is sufficient for your application, with fchmod.
fileCreatefd = mkstemp(fileName);
fchmod(fileCreatefd, 0666)
The mkstemp() function generates a unique temporary filename from template, creates and opens the file, and returns an open file descriptor for the file.
The last six characters of template must be "XXXXXX" and these are replaced with a string that makes the filename unique. Since it will be modified, template must not be a string constant, but should be declared as a character array.
The file is created with permissions 0600, that is, read plus write for owner only. (In glibc versions 2.06 and earlier, the file is created with permissions 0666, that is, read and writefor all users.) The returned file descriptor provides both read and write access to the file. The file is opened with the open(2) O_EXCL flag, guaranteeing that the caller is the process that creates the file.
More generally, the POSIX specification of mkstemp() does not say anything about file modes, so the application should make sure its file mode creation mask (umask(2)) is set appropriately before calling mkstemp() (and mkostemp()).
So after creating the File Use fchmod to change the file permission.