I've read in "The Go Programming Language" that a "given key can be retrieved ... using a constant number of key comparisons on average, no matter how large the hash table." I'm not sure what that means in terms of its implementation internally though. Does that mean it searches through every key until it finds a match or is some type of binary (or other) search algorithm used internally?
For example, if I have a map with 2,000 keys, does it "on average" need to look at 1,000 to find a match or does it look at only 11 (log2 n) as it would with binary search?
Thanks, Ben
The native map type uses a hash table implementation. It uses a hashing function on the key to generate an index into an array of data. Thus, generally, most actions occur in O(1) time. This is only generally true as some keys can result in the same index when hashed, called a collision, which then must be handled specially.
Hash tables are cool!
Maps are implemented as hash tables. There are plenty of places that explain hashing; Here's a nice visualization you can run.
One nice feature of Go is that
From the source file for hashmap:
// A map is just a hash table. The data is arranged
// into an array of buckets. Each bucket contains up to
// 8 key/value pairs. The low-order bits of the hash are
// used to select a bucket. Each bucket contains a few
// high-order bits of each hash to distinguish the entries
// within a single bucket.
//
// If more than 8 keys hash to a bucket, we chain on
// extra buckets.
https://github.com/golang/go/blob/master/src/runtime/map.go
One thing that you can learn from this code that is not normally covered in a lot of classes is how not to invalidate iterators when the map is resized internally.