Open directory using C

Question

I am accepting the path through command line input.

When I do

dir=opendir(args[1]);

it doesn\' t enter the loop...i.e dir==nu

Answer 1

You should really post your code^(a), but here goes. Start with something like:

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

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

        // Ensure correct argument count.

        if (argc != 2) {
            printf ("Usage: testprog <dirname>\n");
            return 1;
        }

        // Ensure we can open directory.

        pDir = opendir (argv[1]);
        if (pDir == NULL) {
            printf ("Cannot open directory '%s'\n", argv[1]);
            return 1;
        }

        // Process each entry.

        while ((pDirent = readdir(pDir)) != NULL) {
            printf ("[%s]\n", pDirent->d_name);
        }

        // Close directory and exit.

        closedir (pDir);
        return 0;
    }

You need to check in your case that args[1] is both set and refers to an actual directory. A sample run, with tmp is a subdirectory off my current directory but you can use any valid directory, gives me: testprog tmp

[.]
[..]
[file1.txt]
[file1_file1.txt]
[file2.avi]
[file2_file2.avi]
[file3.b.txt]
[file3_file3.b.txt]

Note also that you have to pass a directory in, not a file. When I execute:

testprog tmp/file1.txt

I get:

Cannot open directory 'tmp/file1.txt'

That's because it's a file rather than a directory (though, if you're sneaky, you can attempt to use diropen(dirname(argv[1])) if the initial diropen fails).

---

^(a) This has now been rectified but, since this answer has been accepted, I'm going to assume it was the issue of whatever you were passing in.

Answer 2

Here is a simple way to implement ls command using c. To run use for example ./xls /tmp

    #include<stdio.h>
    #include <dirent.h>
    void main(int argc,char *argv[])
    {
   DIR *dir;
   struct dirent *dent;
   dir = opendir(argv[1]);   

   if(dir!=NULL)
      {
   while((dent=readdir(dir))!=NULL)
                    {
        if((strcmp(dent->d_name,".")==0 || strcmp(dent->d_name,"..")==0 || (*dent->d_name) == '.' ))
            {
            }
       else
              {
        printf(dent->d_name);
        printf("\n");
              }
                    }
       }
       close(dir);
     }

Answer 3

Parameters passed to the C program executable is nothing but an array of string(or character pointer),so memory would have been already allocated for these input parameter before your program access these parameters,so no need to allocate buffer,and that way you can avoid error handling code in your program as well(Reduce chances of segfault :)).

Answer 4

Some feedback on the segment of code, though for the most part, it should work...

void main(int c,char **args)

• int main - the standard defines main as returning an int.
• c and args are typically named argc and argv, respectfully, but you are allowed to name them anything

...

{
DIR *dir;
struct dirent *dent;
char buffer[50];
strcpy(buffer,args[1]);

• You have a buffer overflow here: If args[1] is longer than 50 bytes, buffer will not be able to hold it, and you will write to memory that you shouldn't. There's no reason I can see to copy the buffer here, so you can sidestep these issues by just not using strcpy...

...

dir=opendir(buffer);   //this part

If this returning NULL, it can be for a few reasons:

• The directory didn't exist. (Did you type it right? Did it have a space in it, and you typed ./your_program my directory, which will fail, because it tries to opendir("my"))
• You lack permissions to the directory
• There's insufficient memory. (This is unlikely.)