Question

private void SimpleLambda()
{
  dynamic showMessage =  x => MessageBox.Show(x);

  showMessage("Hello World!");
}

错误讯息是：无法将lambda表达式转换为动态类型，因为它不是委托类型

任何帮助，

Answer 1

这与MessageBox无关 - 正如错误消息所示，您根本无法将lambda表达式转换为dynamic，因为编译器不知道创建实例的委托类型是什么

你想：

Action<string> action = x => MessageBox.Show(x);

甚至可以使用方法组转换，但必须匹配返回类型：

Func<string, DialogResult> func = MessageBox.Show;

如果您愿意，可以然后使用dynamic：

dynamic showMessage = action; // Or func
showMessage("Hello World!");

或者，您可以在显式委托实例表达式中指定lambda表达式：

dynamic showMessage = new Action<string>(x => MessageBox.Show(x));

Answer 2

private void SimpleLambda()
{
  Action<string> showMessage =  x => MessageBox.Show(x);

  showMessage("Hello World!");
}

Answer 3

您必须声明委托类型。否则它将不知道它应该是什么类型的lambda表达式 - x可以是任何东西。这应该有效：

Action<string> showMessage = x => MessageBox.Show(x);

有关此委托类型的说明，请参阅Action<T>。

Answer 4

我为此创建了一个类型推理助手。如果我想将它们存储在临时变量中，我不喜欢输入lambdas的签名，所以我写了

var fn = Func.F( (string x) => MessageBox.Show(x) );

或

var fn = Func.F( (double x, double y) => x + y );

你仍然需要输入参数signiture，但是你让类型推断处理返回类型。

实施

using System;

namespace System
{
    /// <summary>
    /// Make type inference in C# work harder for you. Normally when
    /// you want to declare an inline function you have to type
    /// 
    ///     Func<double, double, double> fn = (a,b)=>a+b
    /// 
    /// which sux! With the below methods we can write
    /// 
    ///     var fn = Func.F((double a, double b)=>a+b);
    ///
    /// which is a little better. Not as good as F# type
    /// inference as you still have to declare the args
    /// of the function but not the return value which
    /// is sometimes not obvious straight up. Ideally
    /// C# would provide us with a keyword fun used like
    /// 
    ///     fun fn = (double a, double b)=>a+b;
    ///
    /// but till then this snippet will make it easier
    /// 
    /// </summary>
    public static class Func
    {
        public static Func<A> F<A>(Func<A> f)
        {
            return f; 
        }
        public static Func<A,B> F<A, B>(Func<A, B> f)
        {
            return f; 
        }
        public static Func<A,B,C> F<A, B,C>(Func<A, B,C> f)
        {
            return f; 
        }
        public static Func<A,B,C,D> F<A,B,C,D>(Func<A,B,C,D> f)
        {
            return f; 
        }
        public static Func<A,B,C,D,E> F<A,B,C,D,E>(Func<A,B,C,D,E> f)
        {
            return f; 
        }

        public static Action A(Action f)
        {
            return f; 
        }
        public static Action<_A> A<_A>(Action<_A> f)
        {
            return f; 
        }
        public static Action<_A,B> A<_A, B>(Action<_A, B> f)
        {
            return f; 
        }
        public static Action<_A,B,C> A<_A, B,C>(Action<_A, B,C> f)
        {
            return f; 
        }
        public static Action<_A,B,C,D> A<_A,B,C,D>(Action<_A,B,C,D> f)
        {
            return f; 
        }
        public static Action<_A,B,C,D,E> A<_A,B,C,D,E>(Action<_A,B,C,D,E> f)
        {
            return f; 
        }
    }

}

C＃中的lambda表达式和Messagebox

4 个答案: