CA1013: Overload operator equals on overloading add and subtract

Applies to: Visual Studio Visual Studio for Mac

Note

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Item	Value
RuleId	CA1013
Category	Microsoft.Design
Breaking change	Non-breaking

Cause

A public or protected type implements the addition or subtraction operators without implementing the equality operator.

Rule description

When instances of a type can be combined by using operations such as addition and subtraction, you should almost always define equality to return true for any two instances that have the same constituent values.

You cannot use the default equality operator in an overloaded implementation of the equality operator. Doing so will cause a stack overflow. To implement the equality operator, use the Object.Equals method in your implementation. See the following example.

If (Object.ReferenceEquals(left, Nothing)) Then
    Return Object.ReferenceEquals(right, Nothing)
Else
    Return left.Equals(right)
End If

if (Object.ReferenceEquals(left, null))
    return Object.ReferenceEquals(right, null);
return left.Equals(right);

How to fix violations

To fix a violation of this rule, implement the equality operator so that it is mathematically consistent with the addition and subtraction operators.

When to suppress warnings

It is safe to suppress a warning from this rule when the default implementation of the equality operator provides the correct behavior for the type.

Example 1

The following example defines a type (BadAddableType) that violates this rule. This type should implement the equality operator to make any two instances that have the same field values test true for equality. The type GoodAddableType shows the corrected implementation. Note that this type also implements the inequality operator and overrides Equals to satisfy other rules. A complete implementation would also implement GetHashCode.

using System;

namespace DesignLibrary
{
   public class BadAddableType
   {
      private int a, b;
      public BadAddableType(int a, int b)
      {
         this.a = a;
         this.b = b;
      }
      // Violates rule: OverrideOperatorEqualsOnOverridingAddAndSubtract.
      public static BadAddableType operator +(BadAddableType a, BadAddableType b)
      {
         return new BadAddableType(a.a + b.a, a.b + b.b);
      }
      // Violates rule: OverrideOperatorEqualsOnOverridingAddAndSubtract.
      public static BadAddableType operator -(BadAddableType a, BadAddableType b)
      {
         return new BadAddableType(a.a - b.a, a.b - b.b);
      }
      public override string ToString()
      {
         return String.Format("{{{0},{1}}}", a, b);
      }
   }

   public class GoodAddableType
   {
      private int a, b;
      public GoodAddableType(int a, int b)
      {
         this.a = a;
         this.b = b;
      }
      // Satisfies rule: OverrideOperatorEqualsOnOverridingAddAndSubtract.
      public static bool operator ==(GoodAddableType a, GoodAddableType b)
      {
         return (a.a == b.a && a.b == b.b);
      }

      // If you implement ==, you must implement !=.
      public static bool operator !=(GoodAddableType a, GoodAddableType b)
      {
         return !(a==b);
      }

      // Equals should be consistent with operator ==.
      public override bool Equals(Object obj)
      {
         GoodAddableType good = obj as GoodAddableType;
         if (obj == null)
            return false;
         
        return this == good;
      }

      public static GoodAddableType operator +(GoodAddableType a, GoodAddableType b)
      {
         return new GoodAddableType(a.a + b.a, a.b + b.b);
      }
     
      public static GoodAddableType operator -(GoodAddableType a, GoodAddableType b)
      {
         return new GoodAddableType(a.a - b.a, a.b - b.b);
      }
      public override string ToString()
      {
         return String.Format("{{{0},{1}}}", a, b);
      }
   }
}

Example 2

The following example tests for equality by using instances of the types that were previously defined in this topic to illustrate the default and correct behavior for the equality operator.

using System;

namespace DesignLibrary
{
    public class TestAddableTypes
    {
       public static void Main()
       {
          BadAddableType a = new BadAddableType(2,2);
          BadAddableType b = new BadAddableType(2,2);
          BadAddableType x = new BadAddableType(9,9);
          GoodAddableType c = new GoodAddableType(3,3);
          GoodAddableType d = new GoodAddableType(3,3);
          GoodAddableType y = new GoodAddableType(9,9);
    
          Console.WriteLine("Bad type:  {0} {1} are equal? {2}", a,b, a.Equals(b)? "Yes":"No");
          Console.WriteLine("Good type: {0} {1} are equal? {2}", c,d, c.Equals(d)? "Yes":"No");
          Console.WriteLine("Good type: {0} {1} are == ?   {2}", c,d, c==d? "Yes":"No");
          Console.WriteLine("Bad type:  {0} {1} are equal? {2}", a,x, a.Equals(x)? "Yes":"No");
          Console.WriteLine("Good type: {0} {1} are == ?   {2}", c,y, c==y? "Yes":"No");
       }
    }
}

This example produces the following output:

Bad type:  {2,2} {2,2} are equal? No
Good type: {3,3} {3,3} are equal? Yes
Good type: {3,3} {3,3} are == ?   Yes
Bad type:  {2,2} {9,9} are equal? No
Good type: {3,3} {9,9} are == ?   No

See also

• Equality Operators