Response
If by enough, you mean whether or not your function is injective, the answer is No. The reasoning is that the set of all hash values your function can output has cardinality 2^64, while the space of your inputs is much larger. However, this is not really important, because you can't have an injective hash function given the nature of your inputs. A good hash function has these qualities:
- It's not easily invertible. Given the output k, it's not computationally feasible within the lifetime of the universe to find m such that h(m) = k.
- The range is uniformly distributed over the output space.
- It's hard to find two inputs m and m' such that h(m) = h(m')
Of course, the extents of these really depend on whether you want something that's cryptographically secure, or you want to take some arbitrary chunk of data and just send it some arbitrary 64-bit integer. If you want something cryptographically secure, writing it yourself is not a good idea. In that case, you'd also need the guarantee that the function is sensitive to small changes in the input. The std::hash function object is not required to be cryptographically secure. It exists for use cases isomorphic to hash tables. CPP Rerefence says:
For two different parameters k1 and k2 that are not equal, the probability that std::hash<Key>()(k1) == std::hash<Key>()(k2) should be very small, approaching 1.0/std::numeric_limits<size_t>::max().
I'll show below how your current solution doesn't really guarantee this.
Collisions
I'll give you a few of my observations on a variant of your solution (I don't know what your _class member is).
std::size_t hash_code(const std::unordered_map<std::string, std::string>& m) {
std::hash<std::string> h;
std::size_t result = 0;
for (auto&& p : m) {
result ^= h(p.first) ^ h(p.second);
}
return result;
}
It's easy to generate collisions. Consider the following maps:
std::unordered_map<std::string, std::string> container0;
std::unordered_map<std::string, std::string> container1;
container0["123"] = "456";
container1["456"] = "123";
std::cout << hash_code(container0) << '\n';
std::cout << hash_code(container1) << '\n';
On my machine, compiling with g++ 4.9.1, this outputs:
1225586629984767119
1225586629984767119
The question as to whether this matters or not arises. What's relevant is how often you're going to have maps where keys and values are reversed. These collisions will occur between any two maps in which the sets of keys and values are the same.
Order of Iteration
Two unordered_map instances having exactly the same key-value pairs will not necessarily have the same order of iteration. CPP Rerefence says:
For two parameters k1 and k2 that are equal, std::hash<Key>()(k1) == std::hash<Key>()(k2).
This is a trivial requirement for a hash function. Your solution avoids this because the order of iteration doesn't matter since XOR is commutative.
A Possible Solution
If you don't need something that's cryptographically secure, you can modify your solution slightly to kill the symmetry. This approach is okay in practice for hash tables and the like. This solution is also independent of the fact that order in an unordered_map is undefined. It uses the same property your solution used (Commutativity of XOR).
std::size_t hash_code(const std::unordered_map<std::string, std::string>& m) {
const std::size_t prime = 19937;
std::hash<std::string> h;
std::size_t result = 0;
for (auto&& p : m) {
result ^= prime*h(p.first) + h(p.second);
}
return result;
}
All you need in a hash function in this case is a way to map a key-value pair to an arbitrary good hash value, and a way to combine the hashes of the key-value pairs using a commutative operation. That way, order does not matter. In the example hash_code I wrote, the key-value pair hash value is just a linear combination of the hash of the key and the hash of the value. You can construct something a bit more intricate, but there's no need for that.
