Ger*_*nck 21
我想补充Natrium的答案,即泛型集合也支持这种.ToArray()方法.
List<string> stringList = new List<string>();
stringList.Add("1");
stringList.Add("2");
stringList.Add("3");
string[] stringArray = stringList.ToArray();
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好吧,数组初始化每次都让我得到。所以我花了 10 分钟才把这件事做好。
static void Main(string[] args)
{
String[] as1 = new String[] { "Static", "with", "initializer" };
ShowArray("as1", as1);
String[] as2 = new String[5];
as2[0] = "Static";
as2[2] = "with";
as2[3] = "initial";
as2[4] = "size";
ShowArray("as2", as2);
ArrayList al3 = new ArrayList();
al3.Add("Dynamic");
al3.Add("using");
al3.Add("ArrayList");
//wow! this is harder than it should be
String[] as3 = (String[])al3.ToArray(typeof(string));
ShowArray("as3", as3);
List<string> gl4 = new List<string>();
gl4.Add("Dynamic");
gl4.Add("using");
gl4.Add("generic");
gl4.Add("list");
//ahhhhhh generic lubberlyness :)
String[] as4 = gl4.ToArray();
ShowArray("as4", as4);
}
private static void ShowArray(string msg, string[] x)
{
Console.WriteLine(msg);
for(int i=0;i<x.Length;i++)
{
Console.WriteLine("item({0})={1}",i,x[i]);
}
}
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您还可以new像使用其他对象类型一样使用该运算符:
int[] array = new int[5];
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或者,使用变量:
int[] array = new int[someLength];
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