本文介绍了非法参数异常:n必须为正的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
主类:
public class ECONAPP2 {
static Scanner input= new Scanner(System.in);
static int score = 0;
static ArrayList<Integer> usedArray = new ArrayList<Integer>();
public static void main(String[] args){
app();
arrayContents();
}
public static void arrayContents() {
usedArray.add(2);
usedArray.add(1);
}
应用()方法:
public static void app() {
Random generator = new Random ();
int randomNumber = generator.nextInt(usedArray.size());
System.out.println(randomNumber);
if (randomNumber == 2) {
score();
question2();
usedArray.remove(2);
app();
}
if (randomNumber == 1) {
score();
question1();
usedArray.remove(1);
app();
}
收到此错误:
Exception in thread "main" java.lang.IllegalArgumentException: n must be positive
at java.util.Random.nextInt(Random.java:250)
at ECONAPP2.app(ECONAPP2.java:65)
at ECONAPP2.main(ECONAPP2.java:10)
想不通这是什么意思,什么n为重的presentative?
can't work out what this means and what n is representative of ?
推荐答案
在此行
int randomNumber = generator.nextInt(usedArray.size());
您正试图生成随机数。
不过你有空usedArray,所以返回0。你不能在0范围内产生随机数为0 EXLUSIVE,所以它抛出异常。该值必须是1或更高。
However you have empty usedArray, so it returns 0. You cant generate random number in range 0 to 0 exlusive, so it throws exception. The value must be 1 or higher.
请注意文件:值介于0(含)和指定值(不),因此,例如所有呼叫generator.nextInt(1)
返回0, generator.nextInt(2)
返回0或1 ...
Note documentation : "value between 0 (inclusive) and the specified value (exclusive)", so for example generator.nextInt(1)
return 0 on all calls, generator.nextInt(2)
returns 0 or 1...
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