如何用LinQ对自定义对象求和？

Question

首先，我按照本教程创建了我的Money对象：https : //www.codeproject.com/articles/837791/money-pattern

Money totalItems = _invoice.InvoiceDetails
    .Sum(y => y.Amount); // Amount is of type Money

我在上收到编译异常y.Amount：

无法将类型'Money'隐式转换为'long？' 无法将lambda表达式转换为预期的委托类型，因为该块中的某些返回类型不能隐式转换为委托返回类型

我究竟做错了什么？

这是我的Money课程：

public class Money
{
    public decimal Amount { get; private set; }
    public CurrencyCode Currency { get; private set; }

    #region Constructors
    public Money() { }
    public Money(Money amount)
    {
        this.Amount = amount.Amount;
        this.Currency = amount.Currency;
    }
    public Money(decimal amount, CurrencyCode currencyCode)
    {
        this.Amount = amount;
        this.Currency = currencyCode;
    }
    public Money(int amount, CurrencyCode currency)
        : this(Convert.ToDecimal(amount), currency)
    {
    }
    public Money(double amount, CurrencyCode currency)
        : this(Convert.ToDecimal(amount), currency)
    {
    }
    #endregion

    #region Comprasion operators
    public static bool operator ==(Money var1, Money var2)
    {
        if ((object)var1 == null || (object)var2 == null)
            return false;

        if (var1.Currency != var2.Currency) return false;
        return var1.Amount == var2.Amount;
    }


    public static bool operator !=(Money var1, Money var2)
    {
        return !(var1 == var2);
    }

    public static bool operator >(Money var1, Money var2)
    {
        if (var1.Currency != var2.Currency)
            throw new InvalidOperationException("Comprasion between different currencies is not allowed.");

        return var1.Amount > var2.Amount;
    }

    public static bool operator <(Money var1, Money var2)
    {
        if (var1 == var2) return false;

        return !(var1 > var2);
    }

    public static bool operator <=(Money var1, Money var2)
    {
        if (var1 < var2 || var1 == var2) return true;

        return false;
    }

    public static bool operator >=(Money var1, Money var2)
    {
        if (var1 > var2 || var1 == var2) return true;

        return false;
    }
    #endregion

    #region Ariphmetical operations
    public static Money operator +(Money var1, Money var2)
    {
        if (var1.Currency != var2.Currency)
        {
            throw new InvalidCastException("Calculation is using different currencies!");
        }

        return new Money(var1.Amount + var2.Amount, var1.Currency);
    }

    public static Money operator -(Money var1, Money var2)
    {
        if (var1.Currency != var2.Currency)
        {
            throw new InvalidCastException("Calculation is using different currencies!");
        }

        return new Money(var1.Amount - var2.Amount, var1.Currency);
    }

    public static Money operator *(Money var1, Money var2)
    {
        if (var1.Currency != var2.Currency)
        {
            throw new InvalidCastException("Calculation is using different currencies!");
        }

        return new Money(var1.Amount * var2.Amount, var1.Currency);
    }

    public static Money operator /(Money var1, Money var2)
    {
        if (var1.Currency != var2.Currency)
        {
            throw new InvalidCastException("Calculation is using different currencies!");
        }

        return new Money(var1.Amount / var2.Amount, var1.Currency);
    }

    public static Money operator *(decimal var1, Money var2)
    {
        return new Money(var1 * var2.Amount, var2.Currency);
    }

    public static Money operator *(Money var1, decimal var2)
    {
        return new Money(var1.Amount * var2, var1.Currency);
    }

    public static Money operator /(decimal var1, Money var2)
    {
        return new Money(var1 / var2.Amount, var2.Currency);
    }

    public static Money operator /(Money var1, decimal var2)
    {
        return new Money(var1.Amount / var2, var1.Currency);
    }

    public static Money operator *(int var1, Money var2)
    {
        return new Money(var1 * var2.Amount, var2.Currency);
    }

    public static Money operator *(Money var1, int var2)
    {
        return new Money(var1.Amount * var2, var1.Currency);
    }

    public static Money operator /(int var1, Money var2)
    {
        return new Money(var1 / var2.Amount, var2.Currency);
    }

    public static Money operator /(Money var1, int var2)
    {
        return new Money(var1.Amount / var2, var1.Currency);
    }

    public static Money operator *(long var1, Money var2)
    {
        return new Money(var1 * var2.Amount, var2.Currency);
    }

    public static Money operator *(Money var1, long var2)
    {
        return new Money(var1.Amount * var2, var1.Currency);
    }

    public static Money operator /(long var1, Money var2)
    {
        return new Money(var1 / var2.Amount, var2.Currency);
    }

    public static Money operator /(Money var1, long var2)
    {
        return new Money(var1.Amount / var2, var1.Currency);
    }
    #endregion

    public override bool Equals(object obj)
    {
        if (obj == null) return false;

        Money money = obj as Money;
        return (this.Amount == money.Amount && this.Currency == money.Currency);
    }

    public bool Equals(Money money)
    {
        if ((object)money == null) return false;

        return (this.Amount == money.Amount && this.Currency == money.Currency);
    }

    public override int GetHashCode()
    {
        return base.GetHashCode();
    }

    public override string ToString()
    {
        return this.Amount.ToString();
    }
    #endregion

Answer 1

大概InvoiceDetails是包含public Money Amount属性的类的集合，例如：

public class InvoiceDetail
{
    public Money Amount { get; set; }
}

在这种情况下，您可以使用Enumerable.Aggregate()做总和：

var sum = InvoiceDetails.Aggregate(new Money(0, InvoiceDetails.First().Amount.Currency), (s, d) => s + d.Amount);

为了摆脱稍微难看的new Money(0, InvoiceDetails.First().Amount.Currency)表情，您可能想要引入一个Money.Empty不包含任何货币并且可以添加到任何类型的货币的特殊单例。或修改静态运算符以接受的null值Money并执行以下操作：

var sum = InvoiceDetails.Aggregate((Money)null, (s, d) => s + d.Amount);

或者，引入中间体Select()可以使表达更清晰：

var sum = InvoiceDetails.Select(d => d.Amount).Aggregate((s, a) => s + a);

Enumerable.Sum()不起作用的原因是它是为固定的一组可枚举的算术类型定义的。没有Sum()引入了算术运算符重载的任意类型，因为在这种情况下没有通用的接口或类型推断。（请参见是否存在将我的泛型方法限制为数值类型的约束？答案为“否”。）当然，您可以添加自己的版本，Enumerable.Sum()以支持提供自己的算术类型，请参见此答案为一个开始的地方。