C#中值相等的自定义类的最强大的Equals实现

时间:2009-02-20 17:39:21

标签: c# .net class equals

假设我有一个Point2类,我想实现以下Equals:

public override bool Equals ( object obj )

public bool Equals ( Point2 obj )

这来自有效的C#3书:

public override bool Equals ( object obj )
{
    // STEP 1: Check for null
    if ( obj == null )
    {
        return false;
    }

    // STEP 3: equivalent data types
    if ( this.GetType ( ) != obj.GetType ( ) )
    {
        return false;
    }
    return Equals ( ( Point2 ) obj );
}

public bool Equals ( Point2 obj )
{
    // STEP 1: Check for null if nullable (e.g., a reference type)
    if ( obj == null )
    {
        return false;
    }
    // STEP 2: Check for ReferenceEquals if this is a reference type
    if ( ReferenceEquals ( this, obj ) )
    {
        return true;
    }
    // STEP 4: Possibly check for equivalent hash codes
    if ( this.GetHashCode ( ) != obj.GetHashCode ( ) )
    {
        return false;
    }
    // STEP 5: Check base.Equals if base overrides Equals()
    System.Diagnostics.Debug.Assert (
        base.GetType ( ) != typeof ( object ) );

    if ( !base.Equals ( obj ) )
    {
        return false;
    }

    // STEP 6: Compare identifying fields for equality.
    return ( ( this.X.Equals ( obj.X ) ) && ( this.Y.Equals ( obj.Y ) ) );
}

这有点矫枉过正吗?

4 个答案:

答案 0 :(得分:3)

支持与继承层次结构的相等性是棘手的。你需要弄清楚你的意思。你真的需要继承吗?如果不是 - 如果Point2直接从System.Object派生,并且你可以使它密封,生活变得容易一些。在那种情况下,我会使用:

public override bool Equals (object obj)
{
    return Equals(obj as Point2);
}

public bool Equals (Point2 obj)
{
    // STEP 1: Check for null if nullable (e.g., a reference type)
    // Note use of ReferenceEquals in case you overload ==.
    if (object.ReferenceEquals(obj, null))
    {
        return false;
    }

    // STEP 2: Check for ReferenceEquals if this is a reference type
    // Skip this or not? With only two fields to check, it's probably
    // not worth it. If the later checks are costly, it could be.
    if (object.ReferenceEquals( this, obj))
    {
        return true;
    }

    // STEP 4: Possibly check for equivalent hash codes
    // Skipped in this case: would be *less* efficient

    // STEP 5: Check base.Equals if base overrides Equals()
    // Skipped in this case

    // STEP 6: Compare identifying fields for equality.
    // In this case I'm using == instead of Equals for brevity
    // - assuming X and Y are of a type which overloads ==.
    return this.X == obj.X && this.Y == obj.Y;
}

答案 1 :(得分:2)

不是真的 - 你几乎考虑到了所有可能性。如果此代码用于除临时应用程序之外的任何其他内容,则应考虑此方法的好处,因为由于奇怪的对象相等行为导致的逻辑错误很难调试。

答案 2 :(得分:1)

让我觉得你想要什么。整个区块归结为:

“如果它是完全相同的实例,则返回true。如果它们是具有相同X和Y值的单独实例,则返回true。所有其他情况(null,不同类型,不同x / y值)返回false。”

答案 3 :(得分:1)

它的代码肯定比想要为equals方法编写的代码更多。有很多冗余检查,例如检查ReferenceEquals和HashCodes(我知道这些检查是多余的,因为函数中的最后一行检查结构相等性)。专注于简单易读的代码。

public bool Equals(object o)
{
    Point2 p = o as Point2;
    if (o != null)
        return this.X == o.X && this.Y == o.Y;
    else
        return false;
}

由于您的equals方法使用结构相等,因此请确保使用基于您的字段的实现覆盖GetHashCode。