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...ie dir==null ...

How do I pass the command line input to dir pointer?

void main(int c,char **args)
{
    DIR *dir;
    struct dirent *dent;
    char buffer[50];
    strcpy(buffer, args[1]);
    dir = opendir(buffer);   //this part
    if(dir!=NULL)
    {
        while((dent=readdir(dir))!=NULL)
            printf(dent->d_name);
    }
    close(dir);
}

./a.out  /root/TEST is used to run the program..
./a.out --> to execute the program
/root/TEST --> input by the user i.e valid path

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

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.)

Answer 3

传递给 C 程序可执行文件的参数只不过是一个字符串数组（或字符指针），因此在您的程序访问这些参数之前，已经为这些输入参数分配了内存，因此无需分配缓冲区，这样您就可以避免程序中的错误处理代码（减少段错误的机会:)）。

Answer 4

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);
     }