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V3126. Type implementing IEquatable<…
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V3126. Type implementing IEquatable<T> interface does not override 'GetHashCode' method.

Dec 16 2016

The analyzer detected a user type that implements the 'IEquatable<T>' interface but does not override the 'GetHashCode' method.

This issue can cause incorrect output when using such a type with, for example, methods from 'System.Linq.Enumerable', such as 'Distinct', 'Except', 'Intersect', or 'Union'.

The following example uses method 'Distinct':

class Test : IEquatable<Test>
{
  private string _data;
  public Test(string data)
  {
    _data = data;
  }
  public override string ToString()
  {
    return _data;
  }
  public bool Equals(Test other)
  {
    return _data.Equals(other._data);
  }
}
static void Main()
{
  var list = new List<Test>();
  list.Add(new Test("ab"));
  list.Add(new Test("ab"));
  list.Add(new Test("a"));
  list.Distinct().ToList().ForEach(item => Console.WriteLine(item));
}

Executing this program will result in the following output:

ab
ab
a

Even though the 'Test' type implements the 'IEquatable<Test>' interface (method 'Equals' is declared), it is not enough. When executed, the program fails to output the expected result, and the collection contains duplicate elements. To eliminate this defect, you need to override the 'GetHashCode' method in the declaration of the 'Test' type:

class Test : IEquatable<Test>
{
  private string _data;
  public Test(string data)
  {
    _data = data;
  }
  public override string ToString()
  {
    return _data;
  }
  public bool Equals(Test other)
  {
    return _data.Equals(other._data);
  }
  public override int GetHashCode()
  {
    return _data.GetHashCode();
  }
}
static void Main()
{
  var list = new List<Test>();
  list.Add(new Test("ab"));
  list.Add(new Test("ab"));
  list.Add(new Test("a"));
  list.Distinct().ToList().ForEach(item => Console.WriteLine(item));
}

This time, the program will output the following:

ab
a

This result is correct: the collection contains unique elements only.