【发布时间】:2017-08-06 04:59:36
【问题描述】:
我在互联网上有这段代码,但我无法运行它们请告诉我 为什么???
public static void main( String[] args ) {
keyboard[Scanner] = new Scanner(System.in);
System.out.print( "Hello. What is your name? " );
name[String] = keyboard[Scanner].next();
System.out.print( "Hi, " + name[String] + "! How old are you? " );
age[int] = keyboard[Scanner].nextInt();
System.out.println("So you're " + age[int] + ", eh? That's not so old.");
System.out.print( "How much do you weigh, " + name[String] + "? " );
weight[double] = keyboard[Scanner].nextDouble();
System.out.println( weight[double] + "! Better keep that quiet!!" );
System.out.print("Finally, what's your income, " + name[String] + "? " );
income[double] = keyboard[Scanner].nextDouble();
System.out.print( "Hopefully that is " + income[double] + " per hour" );
System.out.println( " and not per year!" );
System.out.print( "Well, thanks for answering my rude questions, " );
System.out.println( name[String] + "." );
}
请帮帮我!!谢谢!!!
【问题讨论】:
-
你得到什么错误?
-
那些“类型引用”甚至不是 Java。你是从哪里弄到这个的?
-
这是一种在 java 中声明变量的新方法吗?您使用的是哪个神圣的 Java 版本?
-
来自“书”(learnjavathehardway.org/book/ex08.html):我编写了一些代码来展示如果每次使用时都必须添加变量的类型,Java 可能会是什么样子。 不要费心输入它;它不会编译,因为 Java(谢天谢地)不能那样工作。
-
但是这本书告诉你,你不应该尝试运行它,因为它不是有效的java代码!它已经告诉你它错了,那你想从我们这里得到什么?
标签: java error-handling compiler-errors