5
private void SimpleLambda()
{
  dynamic showMessage =  x => MessageBox.Show(x);

  showMessage("Hello World!");
}

错误消息是: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,

4

4 回答 4

16

这与错误消息无关MessageBox- 正如错误消息所说,您根本无法将 lambda 表达式转换为,dynamic因为编译器不知道要创建一个实例的委托类型。

你要:

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

甚至可以使用方法组转换,尽管您必须匹配返回类型:

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

然后,您可以根据需要使用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));
于 2013-08-22T13:02:45.373 回答
5
private void SimpleLambda()
{
  Action<string> showMessage =  x => MessageBox.Show(x);

  showMessage("Hello World!");
}
于 2013-08-22T13:02:53.923 回答
4

您必须声明委托类型。否则它不会知道它应该是什么类型的 lambda 表达式——x可能是任何东西。这应该有效:

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

Action<T>有关此委托类型是什么的说明,请参阅。

于 2013-08-22T13:02:39.970 回答
1

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

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

或者

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

您仍然必须输入参数签名,但您让类型推断处理返回类型。

实施是

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; 
        }
    }

}
于 2013-08-22T13:29:07.177 回答