INT，INT>作为Func键&LT数学平等; [英] Mathematical Equality for Func<int, int>

问题描述

我有一个序键入一个实现 ISequence 。 ISequence 工具的IEnumerable<组件> ，其中元素是另一个自定义类型。目前，我的序类型存储生成一个序列作为 Func键1所述的N个项的说明; INT，INT> 。这是方便，因为它可以让我打电话给 NTERM（INT N）用一个简单的调用函数功能并创建序 s的一个lambda。我想避免改变这种做法。

I have a Sequence type that implements ISequence. ISequence implements IEnumerable<Element> where Element is another custom type. Currently, my Sequence type stores the instructions for generating the N-th term of a sequence as a Func<int, int>. This is convenient because it allows me to call NTerm(int n) with a simple call to the Func and to create Sequences with a lambda. I would like to avoid changing that approach.

我要检查两个序列的平等对象基于两个 Func键平等秒。我开始浏览各地，并有使用表达式 s到打破lambda表达式几个帖子和函数功能 S代表平等，但我说的是数学平等

I would like to check equality of two Sequence objects based on equality of the two Funcs. I began browsing around, and there are a few posts that use Expressions to break down lambdas and Funcs for equality, but I'm talking about mathematical equality.

在换句话说， X => 2 * X 应该等于 C => C * 2 ，随着数学表达式，如Math.Pow或更多地参与运营商的任何变化。如果我能得到那个工作，我可以比较序 S代表数学平等。

In other words, x => 2 * x should equal c => c * 2, along with any variations of mathematical expressions, such as Math.Pow or more involved operators. If I can get that to work, I can compare Sequences for mathematical equality.

我试着写我自己的扩展方法：

I tried to write my own extension method:

public static bool MathEquals(this Expression<Func<int, int>> f,
     Expression<Func<int, int>> g)

我M不知道如何从那里。我有一些基本的默认写：

I'm not sure how to go from there. I have some basic defaults written:

if (ReferenceEquals (f, g)) return true;
if (f == null || g == null) return false;
if (f.NodeType != g.NodeType || f.Type != g.Type) return false;

但我需要检查两个lambda表达式的数学平等，即两个 Func键< INT，INT> 秒。可以这样做？有没有人有办法解决吗？我需要改变我的方法用于存储N个项公式？我反对检查输出，因为序列可以等于对一些输出，而不是全部。

but I need to check the mathematical equality of the two lambda expressions, i.e. the two Func<int, int>s. Can this be done? Does anyone have a solution? Do I need to change my approach for storing the N-th term formula? I'm against checking the output because sequences can be equal for some output and not all.

如果我需要张贴任何序列码的，我会的。

If I need to post any of the Sequence code, I will.

更新：我已经决定，以纪念斯科特的答案接受。然而，工作是不完整的。这里看看第二部分。

UPDATE: I've decided to mark Scott's answer as accepted. However, the work is not complete. Check out part two here.

推荐答案

有两个部分你的问题，我怎么掰开一个表达和evaulate它和怎么办我检查两个表达式意思是相同的逻辑操作。

There are two parts to your question, "How do I break apart a expression and evaulate it", and "How do I check that two expressions mean the same logical operation".

我不知道该怎么办了第二部分，但我知道的第一个。

I don't know how to do the 2nd part, but I do know the first.

你将需要做的是建立一个的 ExpressionVisitor 并把它走过每一个的 BinaryExpression 通过覆盖 VisitBinary 和建设的所有操作的列表

What you will need to do is create a ExpressionVisitor and have it walk through each BinaryExpression by overriding VisitBinary and building up a list of all operations.

public class OperationParser : ExpressionVisitor
{
    public OperationParser()
    {
        Expressions = new List<BinaryExpression>();
    }

    public List<BinaryExpression> Expressions { get; private set; }  

    protected override Expression VisitBinary(BinaryExpression b)
    {
        Expressions.Add(b);

        return base.VisitBinary(b);
    }
}

然后，你会怎么做。

Then you would do

    Expression<Func<int, int>> expression1 = (x) => x + 2;
    Expression<Func<int, int>> expression2 = (y) => 2 + y;

    var parser1 = new OperationParser();
    parser1.Visit(expression1);

    var parser2 = new OperationParser();
    parser2.Visit(expression2);

    //TODO: write a way to compare parser1.Expressions to parser2.Expressions to see if they "mean the same thig"

您只需要填写与你的第二个问题

You just need to fill in the TODO with "your second question"

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