[c] How can I check if a directory exists?

How can I check if a directory exists on Linux in C?

You might use stat() and pass it the address of a struct stat, then check its member st_mode for having S_IFDIR set.

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>

...

char d[] = "mydir";

struct stat s = {0};

if (!stat(d, &s))
  printf("'%s' is %sa directory.\n", d, (s.st_mode & S_IFDIR)  : "" ? "not ");
  // (s.st_mode & S_IFDIR) can be replaced with S_ISDIR(s.st_mode)
else
  perror("stat()");

You may also use access in combination with opendir to determine if the directory exists, and, if the name exists, but is not a directory. For example:

/* test that dir exists (1 success, -1 does not exist, -2 not dir) */
int
xis_dir (const char *d)
{
    DIR *dirptr;

    if (access ( d, F_OK ) != -1 ) {
        // file exists
        if ((dirptr = opendir (d)) != NULL) {
            closedir (dirptr); /* d exists and is a directory */
        } else {
            return -2; /* d exists but is not a directory */
        }
    } else {
        return -1;     /* d does not exist */
    }

    return 1;
}

Use the following code to check if a folder exists. It works on both Windows & Linux platforms.

#include <stdio.h>
#include <sys/stat.h>

int main(int argc, char* argv[])
{
    const char* folder;
    //folder = "C:\\Users\\SaMaN\\Desktop\\Ppln";
    folder = "/tmp";
    struct stat sb;

    if (stat(folder, &sb) == 0 && S_ISDIR(sb.st_mode)) {
        printf("YES\n");
    } else {
        printf("NO\n");
    }
}

According to man(2)stat you can use the S_ISDIR macro on the st_mode field:

bool isdir = S_ISDIR(st.st_mode);

Side note, I would recommend using Boost and/or Qt4 to make cross-platform support easier if your software can be viable on other OSs.

The best way is probably trying to open it, using just opendir() for instance.

Note that it's always best to try to use a filesystem resource, and handling any errors occuring because it doesn't exist, rather than just checking and then later trying. There is an obvious race condition in the latter approach.