C fopen fails for write with errno is 2

c, errno, fopen

Solution

All this tells you is that `errno` had the value 2 after your `fopen` call. You don't know that the call failed, because you didn't check whether `dfile == NULL`. If the output was actually written to the file, presumably the `fopen` call succeeded and the `errno` value was left over from some previous call, likely one you didn't make explicitly.

It's entirely possible for a successful call to `fopen` to set `errno` to a non-zero value.

Failing calls can set `errno` to some non-zero value, but successful calls don't set `errno` to 0. To check for errors, you need to

- Set `errno` to 0 before the call;

- Make the call and check the value it returned to see whether it succeeded or failed; and

- Check the value of `errno` after the call -- but only if you know it failed (otherwise the value of `errno` is meaningless).

If `dfile == NULL`, then the `fprintf` call has undefined behavior; it will probably fail.

On the other hand, you say that `dfile` is `NULL`. How do you know that? Your code doesn't check it. (If the `fopen` call really did fail, could the contents of `C:\List.txt` be left over from a previous run of your program?)

What output do you get from this program?

#include <stdio.h>
#include <errno.h>
int main(void) {
    char debugText [256];
    FILE *dfile;

    sprintf (debugText, "C:\\List.txt");
    dfile = fopen( debugText, "w");
    if (dfile == NULL) {
        printf("fopen failed, errno = %d\n", errno);
    }
    else {
        printf("fopen succeeded\n");
    }
    return 0;
}

Problem

Original source