Java args.length 和命令行参数
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args.length and command line arguments
提问by Wang Pei The Dancer
I got confused with how to use args.length, I have coded something here:
我对如何使用 args.length 感到困惑，我在这里编写了一些代码：
public static void main(String[] args) {
int [] a = new int[args.length];
for(int i = 0;i<args.length;i++) {
System.out.print(a[i]);
}
}
The printout is 0 no matter what value I put in command line arguments, I think I probably confused args.length with the args[0], could someone explain? Thank you.
无论我在命令行参数中输入什么值，打印输出都是 0，我想我可能将 args.length 与 args[0] 混淆了，有人可以解释一下吗？谢谢你。
回答by Maroun
intarray is initialized to zero (all members will be zero) by default, see 4.12.5. Initial Values of Variables:
int默认情况下，数组初始化为零（所有成员都为零），参见4.12.5。变量的初始值：
Each class variable, instance variable, or array component is initialized with a default value when it is created ...
For type int, the default value is zero.
每个类变量、实例变量或数组组件在创建时都使用默认值进行初始化......
对于 int 类型，默认值为零。
You're printing the value of the array, hence you're getting 0.
您正在打印数组的值，因此您将获得0.
Did you try to do this?
你试过这样做吗？
for (int i = 0; i < args.length; i++) {
System.out.print(args[i]);
}
argscontains the command linearguments that are passed to the program. args.lengthis the length of the arguments. For example if you run:
args包含传递给程序的命令行参数。args.length是参数的长度。例如，如果您运行：
java myJavaProgram first second
args.lengthwill be 2 and it'll contain ["first", "second"].
args.length将是 2，它将包含["first", "second"].
And the length of the array args is automatically set to 2 (number of your parameters), so there is no need to set the length of args.
并且数组 args 的长度自动设置为 2（您的参数数量），因此无需设置 args 的长度。
回答by Prasad Kharkar
args[0]is the first element of the argsarray.
args.lengthis the length of the array
args[0]是args数组的第一个元素。
args.length是数组的长度
回答by Bohemian
I think you're missing a code that converts Strings to ints:
我认为您缺少将字符串转换为整数的代码：
public static void main(String[] args) {
int [] a = new int[args.length];
// Parse int[] from String[]
for (int i = 0; i < args.length; i++){
a[i] = Interger.parseInt(args[i]);
}
for (int i = 0; i < args.length; i++){
System.out.print(a[i]);
}
}
回答by Guillaume
The aarray you iterate on is not the argsthat contains the actual arguments. You should try:
a您迭代的数组不是args包含实际参数的数组。你应该试试：
public static void main(String[] args) {
for(int i = 0;i<args.length;i++){
System.out.print(args[i]);
}
}
回答by akhil
the arguements you are passing are stored in args array and not your so called array a. by default a properly declared array if not initialized takes the default values of its datatypes. in your case 0.
您传递的参数存储在 args 数组中，而不是您所谓的数组 a 中。默认情况下，如果未初始化，正确声明的数组将采用其数据类型的默认值。在你的情况下 0。
so you do
所以你也是
public static void main(String[] args) {
for(int i = 0;i<args.length;i++){
System.out.print(args[i]);
}
}
or initialize the array a with the args.
或使用参数初始化数组 a。
回答by Renz Mallari
args.length is = 0;
args.length = 0;
if you're looking for this output:
如果您正在寻找此输出：
args[0]:zero
args[1]:one
args[2]:two
args[3]:three
args[0]:0
args[1]:1
args[2]:2
args[3]:3
here is the example...
这是例子...
public static void main(String[] args) {
公共静态无效主（字符串 [] args）{
// array with the array name "arg" String[] arg={"zero","one","two","three"};
// 数组名称为“arg” String[] arg={"zero","one","two","three"};
for( int i=0; i<arg.length ;++i)
{
System.out.println("args["+i+"]:"+arg[i]);
}
}
}
you have to give a length to the array ..
你必须给数组一个长度..
