相关疑难解决方法(0)

我经常看到数组被初始化如下:

String[] array = new String[] { "foo", "bar", "baz" };

但阅读语言基础 - 数组表明,短语法不需要显式实例化构造函数:

或者,您可以使用快捷语法来创建和初始化数组:

 int[] anArray = { 
     100, 200, 300,
     400, 500, 600, 
     700, 800, 900, 1000
 };

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所以,假设这两种初始化方法:

String[] array = new String[] { "foo", "bar", "baz" };
String[] array2 = { "foo", "bar", "baz" };

这些之间有什么区别吗？双方似乎工作一样,在这种情况下我应该承担的第二个隐式调用的new String[],第一个就是一个更详细的方式,还是有更多的它在幕后？

从Java开始很抱歉,如果这对于一个问题太愚蠢了,但我在网上找不到任何相关信息.

考虑:

public proj 3 {

    static string [][]Item;

    public static void main(String [] args){

        Item[][] = {
             {"BH," , "Backhoe,"            , "200.00"},
             {"ER," , "Electric Rake,"      ,  "10.00"},
             {"EL," , "Electric Lawnmower," ,  "15.00"},
             {"TR," , "Trencher"            ,  "35.00"},
             {"MU," , "Mulcher,"            ,  "20.00"},
             {"TS," , "Tree Sprayer,"       ,  "22.00"},
             {"CP," , "Cider Press,"        ,  "30.00"},
             {"PR," , "Pruner,"             ,  "12.00"},
             {"GE," , "Gas Edger,"          ,  "20.00"},
             {"RO," , "Roller,"             ,   "8.00"},

我怎么能这样做,所以我可以用不同的方法调用数组？

我有以下代码

String[] args = {"a", "b", "c"};
method(args);


private void method(String[] args){
    return args;
}

为什么我不能没有错误地执行以下操作？

method({"a", "b", "c"});

这段代码只是为了证明这一点,而不是我使用的实际方法.我想做第二种方法来清理我的代码,并且当我只使用它们一次传递给我的方法时,避免声明十几个不同的数组.

问题的核心是将字符串数组作为方法参数传递的最有效方法是什么.

我试图初始化一个字符串数组,如下所示,但它有一个错误.

public class Account{
    private String[] account;

    public Account()
    {
        account = {"A", "B", "C"};
    }
}

有谁知道为什么它一直在创建错误？