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I am doing an Unix programming exercise to find out the i-node number of a directory and its parent directory, I know that command "ls -ldi" satisfies my requirement but as I said I am doing Unix programming exercise and just want to further understand it so I write the following code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <dirent.h>

int
main(int argc, char *argv[])
{
    DIR *dp;
    struct dirent *dirp;

    if(argc != 2)
        fprintf(stderr, "usage: ex1 directory_name");

    if((dp = opendir(argv[1])) == NULL)
        fprintf(stderr, "can't open %s", argv[1]);

    while((dirp = readdir(dp)) != NULL)
        if(strcmp(dirp->d_name, ".") == 0 || strcmp(dirp->d_name, "..") == 0)
            printf("inode number of %s is %llu\n", dirp->d_name, dirp->d_ino);

    closedir(dp);

    exit(0);
}

I just check all the entries of a given path and compare if the name is dot or dot-dot, if it is, then print out its d_ino.

Here is what I am confused by: if I run "./a.out .", this acts as what "ls -ldi .", meaning they both give the i-node number of the current directory and its parent directory, but if I run "./a.out /" expecting to find out the i-node number of the root directory and its parent directory, then my program print the i-node number 2 for "." and 1 for ".." but command "ls -ldi /." and "ls -ldi /.." both print 2.

So I am wondering why this occurs.

Thanks!

kezhenxu94
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