如何使用LinQ对自定义对象求和？

Question

首先，我按照本教程创建了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y.Amount上收到了编译异常：

  

无法隐式转换类型   ＆＃39;钱＆＃39;长期？＆＃39;不能   将lambda表达式转换为预期的委托类型，因为有些   块中的返回类型不能隐式转换为   委托返回类型

我做错了什么？

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

据推测，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()来完成总和：

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

要摆脱稍微丑陋的new Money(0, InvoiceDetails.First().Amount.Currency)表达式，您可能需要引入一个不包含任何金钱的特殊单身Money.Empty，并且可以添加到任何类型的资金中。或者修改静态运算符以接受null的{​​{1}}值并执行：

Money

或者，引入中间var sum = InvoiceDetails.Aggregate((Money)null, (s, d) => s + d.Amount); 可能会使表达式更清晰：

Select()

Enumerable.Sum()不起作用的原因是它是为一组固定的可枚举算术类型定义的。对于已引入算术运算符重载的任意类型，没有var sum = InvoiceDetails.Select(d => d.Amount).Aggregate((s, a) => s + a); ，因为此类场景没有通用接口或类型推断。 （请参阅Is there a constraint that restricts my generic method to numeric types?，答案是"no"。）当然，您可以添加自己的Sum()版本，该版本支持提供自己算术运算的类型，例如， this answer是一个可以开始的地方。

Answer 2

与@ dbc的回答类似。您需要创建自己的Sum方法的具体示例。这是因为operator是静态的，而不是可接口的。

我实际上会与dbc略有不同，因为

1. Enumerable.First()的使用会导致IEnumerable
的多次迭代 当没有元素时，
2. Enumerable.First()会失败，但会产生一个不好的例外。
3. 代码重用使事情更加清晰。

相反，我更喜欢。

namespace System.Linq //Using System.Linq namespace makes it easier to use
{
    public static class MoneyEnumerable
    {
        public static Money Sum(this IEnumerable<Money> monies)
        {
            return monies.Aggregate((left, right) => left + right);
        }
    }
}

Answer 3

我最近通过以下两种扩展方法解决了类似的问题。

public static Money Sum<T>(this IEnumerable<T> source, Func<T, Money> selector)
    => Sum(source.Select(selector));

public static Money Sum(this IEnumerable<Money> money)
    => money.Aggregate((left, right) => left + right);

Answer 4

您可以添加一个隐式运算符，允许将对象转换为可以求和的类型。

public static implicit operator decimal(Money money) {
    return money != null ? money.Amount : 0;
}