Remaining disk space in linux and the limits of long - c++

Question

I understand that the disk space in linux could be programmatically retrieved using:
// header for statvfs
#include <sys/statvfs.h>
long GetAvailableSpace(const char* path)
{
struct statvfs stat;
if (statvfs(path, &stat) != 0) {
// error happens, just quits here
return -1;
}
// the available size is f_bsize * f_bavail
return stat.f_bsize * stat.f_bavail;
}
int main(int argc, const char *argv[])
{
// assuming input is the directory which one is interested
printf(" The remaining size is %ld \n", GetAvailableSpace(argv[1]));
}
However on checking the file structure for struct statvfs, they are
struct statvfs {
unsigned long f_bsize; /* filesystem block size */
unsigned long f_frsize; /* fragment size */
fsblkcnt_t f_blocks; /* size of fs in f_frsize units */
fsblkcnt_t f_bfree; /* # free blocks */
fsblkcnt_t f_bavail; /* # free blocks for unprivileged users */
fsfilcnt_t f_files; /* # inodes */
fsfilcnt_t f_ffree; /* # free inodes */
fsfilcnt_t f_favail; /* # free inodes for unprivileged users */
unsigned long f_fsid; /* filesystem ID */
unsigned long f_flag; /* mount flags */
unsigned long f_namemax; /* maximum filename length */
};
Theoretically the size of stat.f_bavail, could be also an unsigned long (I suspect). Wouldn't the multiplication of two unsigned long be much more than a long return type would hold?