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A hash consistent with that equality, computed in constant time.

public int GetHashCode()

Returns — int, a hash of the scalar settings plus whether stop words are present at all.

Example — equal options hash alike, which is the contract.

using Lodestar.Text.Vectorization;

var first = new CountVectorizerOptions { StopWords = new HashSet<string> { "the", "and" } };
var second = new CountVectorizerOptions { StopWords = new HashSet<string> { "and", "the" } };

bool consistent = first.Equals(second) && first.GetHashCode() == second.GetHashCode();  // => True

Remarks — the stop-word set contributes only whether it is present, and deliberately not its contents or its count. Its count cannot be used: Equals compares as a set, so ["the", "the"] equals ["the"] while the counts differ, and equal objects are required to hash alike. Hashing the words themselves would mean hashing all of them, order-independently, which is the O(n) this exists to avoid on a type that may carry a 318-word list.

Unequal options are allowed to collide, and two option sets differing only in their stop words will. That is the correct trade for a type used as a dictionary key occasionally and constructed often.

Applies to — net10.0, netstandard2.0.

See also — CountVectorizerOptions, CountVectorizerOptions.Equals, StopWords.