$cc a.c
$./a.out < inpfilename
I want to print inpfilename on stdout. How do I do that ? Thanks for the help in advance...
$cc a.c
$./a.out < inpfilename
I want to print inpfilename on stdout. How do I do that ? Thanks for the help in advance...
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:
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?Only the parent shell is going to know that. The program, a.out is always going to see it as stdin.
As you put it the process that runs a.out has no notion of the file name of the file that provides its' standard input.
The invocation should be:
and parse
argv
inint main( int argc, char* argv[] ) { ... }
or
And get the filename from stdin.
Then in a.c you need to
fopen
that file to read it's content.An
fstat(0,sb)
(0 is stdin file descriptor) will give you details on the input file, size, permissions (called mode) and inode of the device it resides on.Anyway you won't be able to tell its path: as unix inodes have no idea what path they belong to, and technically (see
ln
) they could belong to more than one path.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 usingfstat
to get an inode number for it then walking the file hierarchy similarly tofind / -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.
In fact, it is possible to get filename from procfs, since /proc/*/fd contains symlink to opened files: