Question

构建一个随机数生成器，它接受上边界输入，并由用户生成数字并打印结果。

无论出于何种原因，我的程序（编译）都不会将边界参数识别为第二个参数，并且根据我的参数检查退出。

在理解如何解决这个问题以及我可能缺少的内容时，我们将不胜感激。

谢谢！

import java.util.Random;
import java.util.Scanner;

public class RandNumGen{
    public static void main(String[] args){

// intro
System.out.println("*****************************************");
System.out.println("*Welcome to the Random Number Generator!*");
System.out.println("*    Discussion 7: Michael Stadnicki    *");
System.out.println("*****************************************");

// ask user to input numbers to be generated
System.out.println("Please enter how many random numbers you would like generated!");
Scanner numGen = new Scanner(System.in);
    int randomNum = numGen.nextInt();
System.out.println("We will generate " + randomNum +" random numbers!");

// ask user to input out boundary for the random numbers
System.out.println("Please enter the boundary for our random numbers to be generated!");
Scanner boundGen = new Scanner(System.in);
    int randomBound = boundGen.nextInt();
System.out.println("We will use " + randomBound +" as our boundary for generation!");

// build our random number generator
Random rNum = new Random();

// check our arguments
if (args.length == 2) {
    randomNum = Integer.parseInt(args[0]);
    randomBound = Integer.parseInt(args[1]);
} else {
    System.out.println("You are required to have 2 arguments on the command line!");
    System.exit(0);
}

// print our random numbers generated based on above bounds and numbers
for (int i = 0; i <randomNum; i++) {
    int numOutput = rNum.nextInt(randomBound + 1);
    System.out.println("Your random numbers are: " + numOutput);
}

}

}

Answer 1

尝试仅询问用户输入，不要使用args。并且不要使用两种不同的扫描仪，只需拨打电话即可 numGen.nextInt（）; 两次。

为什么Java程序无法识别第二个命令行参数？

1 个答案: