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C# C lambda表达式尝试显式指定类型参数_C#_Lambda - Fatal编程技术网

C# C lambda表达式尝试显式指定类型参数

C# C lambda表达式尝试显式指定类型参数,c#,lambda,C#,Lambda,我是C语言的新手,不知道如何在lambda表达式中指定参数。我有以下代码: Dictionary<string,string> MyDictionary = some key + some value; var myReultList= MyDictionary.Select(MyMethod).ToList(); var myReult= await Task.WhenAll(myReultList); private async Task<string> MyMe

我是C语言的新手,不知道如何在lambda表达式中指定参数。我有以下代码:

Dictionary<string,string> MyDictionary = some key + some value;

var myReultList= MyDictionary.Select(MyMethod).ToList();
var myReult= await Task.WhenAll(myReultList);

private async Task<string> MyMethod(string arg1, string arg2){
    //do some async work and return value
}
如何将字典键指定为arg1,将字典值指定为arg2

在这段代码中,我在第二行得到错误:

错误CS0411方法Enumerable.SelectIEnumerable的类型参数不能为空 从用法推断。尝试指定类型参数 明确地说

字典的元素是KeyValuePair,因此需要更改MyMethod的参数类型以匹配:

private async Task<string> MyMethod(KeyValuePair<string, string> pair) 
{
        string arg1 = pair.Key;
        string arg2 = pair.Value;
        ...
}
字典的元素是KeyValuePair,因此需要更改MyMethod的参数类型以匹配:

private async Task<string> MyMethod(KeyValuePair<string, string> pair) 
{
        string arg1 = pair.Key;
        string arg2 = pair.Value;
        ...
}
MyDictionary.Selectkvp=>返回MyMethodkvp.Key,kvp.Value.ToListMyDictionary.Selectkvp=>返回MyMethodkvp.Key,kvp.Value.ToList