在 "params" 中包含另一个参数 (C#)

Include another param in "params" (C#)

string.Format 定义的详尽列表为

public static String Format(IFormatProvider provider,String format, params object[] args);
public static String Format(String format, params object[] args);
public static String Format(String format, object arg0, object arg1, object arg2);
public static String Format(String format, object arg0);
public static String Format(String format, object arg0, object arg1);

和下面的代码

using System;

public class Program
{
    public static void Main()
    {
        var myFormat = "(0:'{0}';1:'{1}')";
        var myParams = new object[] {"arg1", "arg2"};       

        var myString = MyTest(myFormat, myParams);

        Console.WriteLine(myString);
    }

    public static string MyTest(string format, params object[] args)
    {
        string myFirstArg = "arg0";
        var result = string.Format(format, myFirstArg, args);
        return result;
    }
}

会给

(0:'arg0';1:'System.Object[]')

现在的问题是如何在 myParams 中包含 arg0,以便输出

(0:'arg0';1:'arg1';2:'arg2')

PS。 (假设myFormat = "(0:'{0}';1:'{1}';2:'{2}')";)

您必须创建另一个以 arg0 作为第一个元素的数组。试试这个:

using System.Linq;
public static string MyTest(string format, params object[] args) {
    var newarr = new object[] { "arg0" };
    newarr.Concat(args);
    var result = string.Format(format, newarr);
    return result;
}

您可以使用 LINQ:

public static string MyTest(string format, params object[] args)
{
    string myFirstArg = "arg0";
    var result = String.Format(format, (new[] { myFirstArg }).Union(args).ToArray());
    return result;
}

我使用 Prepend 扩展方法来完成此类任务:

class Program
{
    static void Main()
    {
        string myFirstArg = "arg0";
        var format = "(0:'{0}';1:'{1}';2:'{2}')";
        var args = new object[] { "arg1", "arg2" };
        var result = string.Format(format, args.Prepend(myFirstArg).ToArray());
    }
}
public static class IEnumerableExtensions
{
    public static IEnumerable<T> Prepend<T>(this IEnumerable<T> collection, T element)
    {
        yield return element;
        foreach (var e in collection) yield return e;
    }
}