Opening a file gives unexpected return value

Question

could someone please explain me, why file opening is not successful? why printing "file" will give -1? Is it a problem with char *source?

int opf(char *source){
    int file;
    file=open(source,O_RWR);
    printf("%d",file); 
}

And is it possible to do something like this: file is in another directory, so

int opf(char *source){
    int file;
    file=open("some_directory/ %s",source,O_RWR);
    printf("%d",file); 
}

here I get the "makes integer from pointer without a cast" error. I tried many different things but i guess the problem lies with me not grasping correctly the concepts.

As for your second and unrelated question (you should really post two questions) search for and read about `snprintf`. — Some programmer dude, May 12 '16 at 08:59
This has nothing to do with pointers; you're trying to use `open` as if it were `printf`. But it's not `printf`. It's `open`. — Lightness Races in Orbit, May 12 '16 at 09:00
You can't make things up and expect them to work or even expect to get a useful error message when you try them. — David Schwartz, May 12 '16 at 09:01

Sourav Ghosh · Accepted Answer · 2016-05-12T09:02:03.480

From the man page of open() (emphasis mine)

Upon successful completion, the function shall open the file and return a non-negative integer representing the lowest numbered unused file descriptor. Otherwise, -1 shall be returned and errno set to indicate the error. No files shall be created or modified if the function returns -1.

So, in your case, the open() has failed. Check for the errno to get more details on that. All the possible values for errno is also documented in the linked page.

That being said,

O_RWR does not seem to be a valid mode,O_RDWR is.
For the second approach, you can use sprintf()/snprintf() first to create the path string and then you can pass that to open().

Opening a file gives unexpected return value

1 Answers1