I'm running an Linux OS and trying to open file in C compiler like this :
file = fopen ("list.txt", "r");
but the file is not opend!
and when i put the full path like this :
file = fopen ("/home/rami/Desktop/netfilter/list.txt", "r");
it is working!
why the first example is not working?
the list.txt is in the same directory of the c file
thanks.
It's not the directory of the C file that matters, it's your current working directory that does. Try
cd /home/rami/Desktop/netfilter
before running the executable.
Do you know WHY the file didn't open?
Always check the return value of fopen() (and most other functions) and report back a readable error.
file = fopen("file.txt", "r");
if (!file) {
perror("file open");
exit(EXIT_FAILURE);
}
I see you've already found out what your problem is, but the above is a suggestion for the future (and to change your current project)
Is the executable also in the same dir as that of list.txt?
Edit: Actually that doesnt matter. It's the current working dir as per the other answer.
Related
I'm new to c programming. I have the following directory structure.
Project
|____ dataset
| |_____ data.txt
|
|____ main.c
My code:
FILE *stream = fopen("dataset/data.txt", "r");
or
FILE *stream = fopen("./dataset/data.txt", "r");
returns this error: No such file or directory
But when I use the absolute path it works without any error:
FILE *stream = fopen("/home/<user_name>/C/<project_name>/dataset/data.txt", "r");
What am I doing wrong here?
Read path_resolution(7).
You may want to use chdir(2), since fopen(3) calls open(2) and all of them could fail.
When fopen fails (or some syscalls(2) fails) use errno(3) or perror(3).
Be aware of glob(3), getcwd(2) and fnmatch(3).
See also proc(5).
I have posted this to document my issue, see my self-answer below.
No matter what I try, fopen(...) cannot open the existing file at the path which exists and returns NULL. I am executing the program from a bash script in ~/path. The program file is stored at ~/path/to.
int main(void) {
const char* filename = "my/file"
FILE* fp = NULL;
fp = fopen(filename, "r"); // file is still NULL, segfaults on indirection
if (!fp) exit(1);
fclose(fp);
}
fopen(3) is documented as capable of failing:
Otherwise, NULL is returned and errno is set to indicate the error.
So you should at least code:
FILE* fp = fopen(filename, "r");
if (fp == NULL) { perror(filename); exit(EXIT_FAILURE); };
and fopen won't even try to create a file that you open for reading only.
As a rule of thumb, you always need to check against failure of fopen (a minima like above), and report to your user (with the help of errno(3), perror(3), strerror(3) -used as strerror(errno)- ...) the reason of that failure. An educated user would be able to manage (perhaps with help from his sysadmin).
ENOENT is documented in errno(3) to mean
ENOENT No such file or directory (POSIX.1-2001).
Typically, this error results when a specified path‐
name does not exist, or one of the components in the
directory prefix of a pathname does not exist, or the
specified pathname is a dangling symbolic link.
I find that explanation pretty clear. In your case, you probably don't have any path/ directory in your current working directory, or you do have path/to/my/ directory without any file entry, etc (e.g. path/ exists but without to/ inside it) ....
You could improve your program by showing not only the errno (using strerror(errno) or perror) but also the working directory. See getcwd(3). Or you could leave your user to guess it. Your user could have changed the working directory, e.g. with a cd builtin command of his unix shell.
I'm trying to open files in a C program, but I am unsure where to place the files I want to open (as in which directory). Here is the code, but I really just need to know where to place the file I want to open with fopen().
FILE *fileptr;
fileptr = fopen("QuizQuestions.txt", "r");
if (fileptr == NULL) {
printf("Unable to open file.");
}
Any help is appreciated!
If you don't use an absolute pathname in your code, paths are interpreted relative to the working directory of the user when they run the program. So for your program, the user should put the file in their current directory.
The location of the program itself is irrelevant. If you want to get the location of the program, you see this question:
How do I find the location of the executable in C?
You can then concatenate the directory with the filename.
You need to keep files where source code file is placed.Otherwise, you need to give absolute path.
I am using C to open a file for reading. I have this code :
fp = fopen("./settings.cfg","r");
if (fp != NULL)
printf("OK");
else
printf("ERROR");
but I always get an error.
The file is located in the folder where the executable resides. I have tried writing only "settings.cfg". What might be the problem?
Try perror() to have the library itself tell you what, if anything, is wrong.
fp = fopen("./settings.cfg", "r");
if (fp != NULL)
printf("OK\n");
else perror("fopen");
You are opening the file in the "current directory", not "in the folder where the executable is".
In fact, unix has no easy way to find that particular folder; in Linux you could readlink() the /proc/[your pid]/exe link to find the executable and strip off the filename portion -- that will work in many cases, but there are some fringe cases like hardlinks that will make it fail.
From which directory do you run the program? It won't have the directory where it resides as the current directory, it will be inherited from the environment.
Could also be rights, that the file is owned by someone else and you don't have read rights.
Also double-check the filename. This sounds obvious, but do it anyway.
If file you are trying to open is in the same directory of your C compiled file, you must simply do
fp = fopen("settings.cfg","r");
if (fp != NULL)
printf("OK");
else
printf("ERROR");
without the initial " ./ " at the file's name
Since C it's not a language I am used to program with, I don't know how to do this.
I have a project folder where I have all the .c and .h files and a conf folder under which there is a config.txt file to read. How can I open that?
FILE* fp = fopen("/conf/config.txt", "r");
if (fp != NULL)
{
//do stuff
}
else
printf("couldn't open file\n");
I keep getting the error message. Why?
Btw, this only have to work on windows, not linux.
Thanks.
The easy way is to use an absolute path...
my_file = fopen("/path/to/my/file.txt", "r");
Or you can use a relative path. If your executable is in /home/me/bin and your txt file is in /home/me/doc, then your relative path might be something like
my_file = fopen("../doc/my_file.txt", "r");
The important thing to remember in relative paths is that it is relative to the current working directory when the executable is run. So if you used the above relative path, but you were in your /tmp directory and you ran /home/me/bin/myprog, it would try to open /tmp/../doc/my_file.txt (or /doc/my_file.txt) which would probably not exist.
The more robust option would be to take the path to the file as an argument to the program, and pass that as the first argument to fopen. The simplest example would be to just use argv[1] from main's parameters, i.e.
int main(int argc, char **argv)
{
FILE *my_file = fopen(argv[1], "r");
/* ... */
return 0;
}
Of course, you'll want to put in error checking to verify that argc > 2, etc.
You're probably going to want to look into the dirent family of routines for directory traversal.
The location of your .c and .h files is not really the issue; the issue is the current working directory when you run your executable.
Can you not pass in the full path to the fopen() function?