what does dividing by sizeof(void *) mean?

Question

I'm working with hash tables and I came across this function. But what does hash / sizeof(void *) mean? and the comment given after it - get rid of known-0 bits?

// This matches when the hashtable key is a pointer.
      template<class HashKey> class hash_munger<HashKey*> {
       public:
        static size_t MungedHash(size_t hash) {
          // TODO(csilvers): consider rotating instead:
          //    static const int shift = (sizeof(void *) == 4) ? 2 : 3;
          //    return (hash << (sizeof(hash) * 8) - shift)) | (hash >> shift);
          // This matters if we ever change sparse/dense_hash_* to compare
          // hashes before comparing actual values.  It's speedy on x86.
          return hash / sizeof(void*);   // get rid of known-0 bits
        }
      };

Answer 1

On most machines and ABIs, pointers are generally word-aligned. So dividing by the sizeof pointers is essentially ignoring the least bits (e.g. 3 lowest bits of bytes addresses are 0 on most 64 bits processors since a 64 bits word has 8 bytes, each of 8 bits).