$cc a.c
$./a.out < inpfilename
I want to print inpfilename on stdout. How do I do that ? Thanks for the help in advance...
You can't get the filename exactly as input; the shell will handle all that redirection stuff without telling you.
In the case of a direct < file
redirection, you can retrieve a filepath associated with stdin by using fstat
to get an inode number for it then walking the file hierarchy similarly to find / -inum
to get a path that matches it. (There might be more than one such filepath due to links.)
But you shouldn't ever need to do this. As others have said, if you need to know filenames you should be taking filenames as arguments.
Why do you want to do this? All your program a.out
is passed from the shell, is an open file descriptor, stdin.
The user might as well do this:
cat inpfilename | ./a.out
and now you have absolutely no filename to use (except /dev/stdin).
If a.out
needs to work with filenames, why not take the file as a command-line argument?
In fact, it is possible to get filename from procfs, since /proc/*/fd contains symlink to opened files:
char filename[bufsize];
int sz = readlink("/proc/self/fd/0", filename, bufsize-1);
filename[sz] = 0;
puts(filename);