检查操作委托回调中的泛型类型

本文关键字:泛型类型 回调 操作 检查 | 更新日期: 2023-09-27 18:32:13

所以这是代码,

public void DoSomething<T>(string key, Action<T> callback)
{
    Type typeParameterType = typeof(T);
    if (typeParameterType.Equals(typeof(string)))
    {
        callback("my string response");
    }
    if (typeParameterType.Equals(typeof(int)))
    {
        callback(1); // my int response
    }
    // etc...
}

但是,我收到错误...我是所有 C# 泛型和委托内容的新手。

我得到的错误是,

Error   1   Delegate 'System.Action<T>' has some invalid arguments
Error   2   Argument 1: cannot convert from 'string' to 'T' 

对我来说,创建简单而习惯的美丽而有用的方法很重要。

所以我很想像这样实现上面的例子,

int total = 0;
DoSomething<int>("doesn't matter", x => {
    total = 10 + x; // i can do this because x is an INT!!! (:
});
string message = "This message is almost ";
DoSomething<int>("doesn't matter", x => {
    message = "finished!!!"; // i can do this because x is an STRING!!! (:
});

但是我被困住了...请帮忙!

=============================================================================================================================================================================================================================================

==

正如达斯布林肯莱特指出的那样,

重载是最干净、编译器友好的方法...我的 API 现在看起来像,

DoSomething("doesn't matter", new Action<string>(x => {
    message = "finished!!!"; // i can do this because x is an STRING!!! (:
}));

这是付出的小代价,更容易理解。

感谢您的回答(:

=============================================================================================================================================================================================================================================

==

再做一些研究,我真的可以通过执行以下操作来清理它;

DoSomething("doesn't matter", (string x) => {
    message = "finished!!!"; // i can do this because x is an STRING!!! (:
});

请注意:(字符串 x)

现在编译器知道了!很酷吧?

检查操作委托回调中的泛型类型

特定类型(如 intstring)不能强制转换为T,但object可以。这应该有效:

if (typeParameterType.Equals(typeof(string)))
{
    callback((T)((object)"my string response"));
}
if (typeParameterType.Equals(typeof(int)))
{
    callback((T)((object)1)); // my int response
}

然而,有点奇怪的是,你首先需要这样做:与其跳过泛型的箍,不如使用多种方法更优雅地处理问题:

public void DoSomething(string key, Action<int> callback) {
    callback(1);
}
public void DoSomething(string key, Action<string> callback) {
    callback("my string response");
}

现在你可以像这样调用这些方法:

DoSomething("hello", new Action<int>(x => Console.WriteLine("int: {0}", x)));
DoSomething("world", new Action<string>(x => Console.WriteLine("str: {0}", x)));

或者像这样:

DoSomething("hello", (int x) => Console.WriteLine("int: {0}", x));
DoSomething("world", (string x) => Console.WriteLine("str: {0}", x));

你可以检查回调类型:

public void DoSomething<T>(string key, Action<T> callback)
{
    var action1 = callback as Action<string>;
    if (action1 != null)
    {
        action1("my string response");
        return;
    }
    var action2 = callback as Action<int>;
    if (action2 != null)
    {
        action2(1); // my int response
        return;
    }
    // etc...
}