由网友(愿你喜)分享简介：可能显示的文件： Enumerable.Cast扩展方法不能投从int长，为什么呢？ Puzzling Enumerable.Cast InvalidCastException的 Cast/Convert IEnumerable的＆LT; T＆GT;到了IEnumerable＆LT; U＆GT; ？ Possi...

可能显示的文件：   Enumerable.Cast<T>扩展方法不能投从int长，为什么呢？   Puzzling Enumerable.Cast InvalidCastException的   Cast/Convert IEnumerable的＆LT; T＆GT;到了IEnumerable＆LT; U＆GT; ？

Possible Duplicates: Enumerable.Cast<T> extension method fails to cast from int to long, why ? Puzzling Enumerable.Cast InvalidCastException Cast/Convert IEnumerable<T> to IEnumerable<U> ?

我想要一个整数数组转换为double数组（这样我就可以把它传递给一个函数，它双打的数组）。

I'm trying to convert an array of integers to an array of doubles (so I can pass it to a function that takes an array of doubles).

最明显的解决方案（对我来说，至少）是使用Cast扩展功能IEnumerable的，但它给了我一个InvalidCastException，我不明白为什么。我的解决方法是使用选择，而不是，但我认为演员看起来更加整洁。

The most obvious solution (to me, at least) is to use the Cast extension function for IEnumerable, but it gives me an InvalidCastException, and I don't understand why. My workaround is to use Select instead, but I think Cast looks neater.

有人能告诉我，为什么铸造法是不工作？

Could someone tell me why the Cast method isn't working?

但愿code以下说明我的问题：

Hopefully the code below illustrates my problem:

namespace ConsoleApplication1
{
    using System;
    using System.Collections.Generic;
    using System.Linq;

    class Program
    {
        static void Main()
        {
            var intArray = new[] { 1, 2, 3, 4 };
            PrintEnumerable(intArray, "intArray: ");

            var doubleArrayWorks = intArray.Select(x => (double)x).ToArray();
            PrintEnumerable(doubleArrayWorks, "doubleArrayWorks: ");

            // Why does this fail??
            var doubleArrayDoesntWork = intArray.Cast<double>().ToArray();
            PrintEnumerable(doubleArrayDoesntWork, "doubleArrayDoesntWork: ");

            // Pause
            Console.ReadLine();
        }

        private static void PrintEnumerable<T>(
            IEnumerable<T> toBePrinted, string msgPrefix)
        {
            Console.WriteLine(
                msgPrefix + string.Join(
                    ",", toBePrinted.Select(x => x.ToString()).ToArray()));
        }
    }

}

推荐答案

这个问题是来自投运营商（重载）的事实，在编译时解决。 试想想怎样投实现。我敢打赌，code是这样的：

The problem comes from the fact that cast operators (overloaded) are resolved at compile time. Try to think how Cast is implemented. I bet the code looks like this:

public static IEnumerable<T> Cast<T>(this IEnumerable source)
{
   foreach(object element in source)
   {
      yield return (T)element;
   }
}

所有的编译器的信息是，它需要对象转换为T类型而对于它会使用默认的继承铸造。没有自定义重载运算符将被使用。而在你的榜样一个int是不是双这样的演员会失败。

All the info the compiler has is that it needs to cast an object to a type T. And for that it will use the default inheritance casting. No custom overloaded operator will be used. And in your example an int is not a double so the cast will fail.

在选择例如：

source.Select(a => (double)a));

工作，因为编译器知道这两种类型，它是能够调用相应的重载运算符。

works because the compiler knows both types and it's able to call the appropriate overloaded operator.