尼斯，干净的交叉只使用扩展方法Linq中加入 [英] Nice, clean cross join in Linq using only extension methods

问题描述

可能重复：结果
的嵌套“从”的LINQ查询与扩展方法

表示

我敢肯定，这个以前被要求，但老实说，我无法找到任何

我很好奇相当于语法会是怎样以下仅使用内置的LINQ扩展方法：

  VAR Z1 = $ b $从X b在XS 
从Ÿ在YS 
选择新的{X，Y }; 
  

我能得到与此相同的结果：

  VAR Z2 = xs.SelectMany（X => ys.Select（Y =>新建{X，Y}））; 
  

但它产生不同的IL代码，代码是有点令人费解，很难理解。 ？是否有一个更清洁的方式使用扩展方法来做到这一点。

---

下面是为写我的整个测试方法：

 私人无效测试（）
 {
变种XS =新[] {1D，2D，3D}; 
变种YS =新[] {4D，5D，6D};从X在XS 
 
 
变种Z1 = 
在YS 
从Ÿ选择新的{X，Y}; 
 
 VAR Z2 = xs.SelectMany（X => ys.Select（Y =>新建{X，Y}））; 
} 
  

这里是：IL代码 XS =新的双[]
{
1.0，
2.0，
3.0
};
双[] =伊苏新的双[]
{
4.0，
5.0，
6.0
};
变种Z = $ B $从X在XS $从Y B $ B在YS
B选择新的
{
X = X，$ B $通过= Y
};
VAR Z2 = xs.SelectMany（（双X）=>从YS
Y形
选择新的
{
X = X，$ B $通过= Y
}）;
}

解决方案

一个办法是：

  VAR Z2 = xs.SelectMany（X => YS，（X，Y）=>新建{X，Y} ）; 
  

Possible Duplicate:
Nested “from” LINQ query expressed with extension methods

I'm sure this has been asked before, but I honestly couldn't find anything.

I'm curious what the equivalent syntax would be for the following using only built-in Linq extension methods:

var z1 =
    from x in xs
    from y in ys
    select new { x, y };

I can get the same results with this:

var z2 = xs.SelectMany(x => ys.Select(y => new { x, y }));

But it produces different IL code, and the code is a bit convoluted and hard to understand. Is there a cleaner way to do this with extension methods?

---

Here's my entire test method as written:

private void Test()
{
    var xs = new[] { 1D, 2D, 3D };
    var ys = new[] { 4D, 5D, 6D };

    var z1 =
        from x in xs
        from y in ys
        select new { x, y };

    var z2 = xs.SelectMany(x => ys.Select(y => new { x, y }));
}

Here's the [Edit: C# interp of the] IL code (using ILSpy):

private void Test()
{
    double[] xs = new double[]
    {
        1.0, 
        2.0, 
        3.0
    };
    double[] ys = new double[]
    {
        4.0, 
        5.0, 
        6.0
    };
    var z =
        from x in xs
        from y in ys
        select new
        {
            x = x,
            y = y
        };
    var z2 = xs.SelectMany((double x) =>
        from y in ys
        select new
        {
            x = x,
            y = y
        });
}

解决方案

One way would be:

var z2 = xs.SelectMany(x => ys, (x, y) => new {x, y});

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