是否有任何理由不使用.NET中的GetHashCode生成条件哈希码？

Question

In a class I use as a document or document+page identifier, I use the following implementation of GetHashCode. It 'felt' right but since I haven't really seen domain-specific conditioning in this method before, I was wondering if there is a reason not to do so. Of course the ToString method has the same conditioning as well.

public override int GetHashCode ()
{
    int hash = 0;

    unchecked
    {
        hash = 17;
        hash = hash * 23 + this.ProductManufacturer.Value.GetHashCode();
        hash = hash * 23 + this.ProductName.Value.GetHashCode();
        hash = hash * 23 + this.ProductVersion.Value.GetHashCode();
        hash = hash * 23 + this.Guid.Value.GetHashCode();

        if (this.Type != IdentifierType.Document)
        {
            hash = hash * 23 + this.PageNumber.Value.GetHashCode();
            hash = hash * 23 + this.PageCount.Value.GetHashCode();
        }
    }

    return (hash);
}

Updated Code based on answers and Eric's links :

public bool Equals (Identifier other)
{
    return (this.Equals(other, this.Type));
}

public override bool Equals (object obj)
{
    return ((obj is HouseOIdentifier) && (this.Equals(obj as Identifier)));
}

public bool Equals (Identifier other, IdentifierType type)
{
    bool result = false;

    if (object.ReferenceEquals(this, other))
    {
        result = true;
    }
    else if (!object.ReferenceEquals(other, null))
    {
        result
            = (this.Type == other.Type)
            && (this.ProductManufacturer.Key == other.ProductManufacturer.Key)
            && (this.ProductManufacturer.Value == other.ProductManufacturer.Value)
            && (this.ProductName.Key == other.ProductName.Key)
            && (this.ProductName.Value == other.ProductName.Value)
            && (this.ProductVersion.Key == other.ProductVersion.Key)
            && (this.ProductVersion.Value == other.ProductVersion.Value)
            && (this.Guid.Key == other.Guid.Key)
            && (this.Guid.Value == other.Guid.Value)
            ;

        if (type == IdentifierType.Page)
        {
            result
                &= (this.PageNumber.Key == other.PageNumber.Key)
                && (this.PageNumber.Value == other.PageNumber.Value)
                && (this.PageCount.Key == other.PageCount.Key)
                && (this.PageCount.Value == other.PageCount.Value)
                ;
        }
    }

    return (result);
}

public override int GetHashCode ()
{
    int hash = 0;

    unchecked // Overflow is fine, just wrap.
    {
        hash = 17;
        hash = hash * 23 + this.Type.GetHashCode();
        hash = hash * 23 + this.ProductManufacturer.Key.GetHashCode();
        hash = hash * 23 + this.ProductManufacturer.Value.GetHashCode();
        hash = hash * 23 + this.ProductName.Key.GetHashCode();
        hash = hash * 23 + this.ProductName.Value.GetHashCode();
        hash = hash * 23 + this.ProductVersion.Key.GetHashCode();
        hash = hash * 23 + this.ProductVersion.Value.GetHashCode();
        hash = hash * 23 + this.Guid.Key.GetHashCode();
        hash = hash * 23 + this.Guid.Value.GetHashCode();

        if (this.Type == HouseOfSynergy.FastForm.Core.Identifier.EnumType.Page)
        {
            hash = hash * 23 + this.PageNumber.Key.GetHashCode();
            hash = hash * 23 + this.PageNumber.Value.GetHashCode();
            hash = hash * 23 + this.PageCount.Key.GetHashCode();
            hash = hash * 23 + this.PageCount.Value.GetHashCode();
        }
    }

    return (hash);
}

public override string ToString ()
{
    return ("whatever");
}

Answer 1

Your implementation is fine as long as equal objects have the same hash code. That's the only requirement there is. How you calculate the hash code, that is, whether you use if-statements or not, is not important.

(Also, it would be nice if the hash code never changed during the object's life time, because it breaks pretty much any data structure that uses hash codes. Optimally the object should be immutable.)