Question

我有这种结构 - 这里是一个虚拟实现：

interface IBasicWaveform
{
    double FindAmplitudeFactor(int samplesPerSecond);
    double ValueAtPhase(double phase);
}

abstract class BaseWaveform : IBasicWaveform
{
    public double FindAmplitudeFactor(int samplesPerSecond)
    {
        throw new NotImplementedException();
    }

    public double ValueAtPhase(double phase)
    {
        throw new NotImplementedException();
    }
}

sealed class SawDownWaveform : BaseWaveform
{
}

sealed class SawUpWaveform : BaseWaveform
{
}

我想创建一个带有Description（字符串）和BaseWaveform DATA TYPE作为TValue的Dictionary

所以我想写这样的代码：

    var myDict = new Dictionary<string,  BaseWaveform**Datatype** >()
    {
        {"Saw down", SawDownWaveform},
        {"Saw up", SawUpWaveform}
    }

我也试过这个：

    var myDict = new Dictionary<string, Type>()    
    {
        {"Saw down", (Type)SawDownWaveform},
        {"Saw up", (Type)SawUpWaveform}
    };

但它给了我一个编译错误：

'WindowsFormsApplication5.SawDownWaveform'是'type'，但用作'变量'

问题是：我怎样才能在c＃中做到这一点？

Answer 1

typeof运算符是在编译时给出类型标识符时获取Type对象的方式：

var myDict = new Dictionary<string, Type>()    
{
    {"Saw down", typeof(SawDownWaveform)},
    {"Saw up", typeof(SawUpWaveform)}
};

Answer 2

我相信您只是想使用typeof()运算符而不是尝试转换为(Type)。

var myDict = new Dictionary<string, Type>()    
{
    {"Saw down", typeof(SawDownWaveform)},
    {"Saw up", typeof(SawUpWaveform)}
};

使用数据类型作为字典中的值

2 个答案: