Question

I’m trying to figure out how exactly to use stat() to capture information about a file. What I need is to be able to print several fields of information about a file. So..

 #include <iostream>
 #include <sys/types.h>
 #include <sys/stat.h>
 #include <fcntl.h>
 using namespace std;

 int main() {
     struct stat buf;
     stat("file",&buf);
               ...
     cout << st_dev << endl;
     cout << st_ino << endl;
     cout << st_mode << endl;
     cout << st_nlink << endl;
     cout << st_uid << endl;
     cout << st_gid << endl;
     cout << st_rdev << endl;
     cout << st_size << endl;
     cout << st_blksize << endl;
     cout << st_blocks << endl;
     cout << st_atime << endl;
     cout << st_mtime << endl;
     cout << st_ctime << endl;
     ...
 }

I’m thoroughly confused about how to do this. Why is &buf a parameter to stat? I don’t care about storing this information in memory, I just need the outputted fields within my c++ program. How do I access the information contained in the struct? Is buf actually supposed to contain the returned information from stat()?

Answer 1

Yes, buf is being used here as an out-parameter. The results are stored in buf and the return value of stat is an error code indicating if the stat operation succeeded or failed.

It is done this way because stat is a POSIX function, designed for C, which does not support out-of-band error reporting mechanisms like exceptions. If stat returned a struct, then it would have no way to indicate errors. Using this out-parameter method also allows the caller to choose where they want to store the results, but that’s a secondary feature. It’s perfectly fine to pass the address of a normal local variable, just like you have done here.

You access the fields of a struct like you would any other object. I presume you are at least familar with object notation? E.g. the st_dev field within the stat struct called buf is accessed by buf.st_dev. So:

cout << buf.st_dev << endl;

etc.