使用随机数填充ArrayList,然后使用Print Array填充

时间:2017-03-05 00:54:58

标签: java arrays arraylist

我想用随机数填充arrayList然后打印数组。但是,在执行程序时出现了大量错误。任何帮助将不胜感激。

public class methods {
//variables
int capacity;
private static ArrayList<Double> randomArray;



public methods(int capacity) {
  //default constructor to initalize variables and call populateArray to
  //populate ArrayList with random numbers
  randomArray = new ArrayList<>(capacity); 
  populateArray();
}

 //Method that populates Array with random numbers
private void populateArray()
{

   Random rand = new Random();

   for (int i=0; i<= capacity; i++)
   {
       double r = rand.nextInt() % 256;

       randomArray.add(i,r);

   }

}
 //Get Array adds numbers to the string that is called in my main class and printed
public String getArray() {
String result = "";
for (int i=0; i<= capacity; i++)
   {
       result += String.format("%4d", randomArray);

   }
return result;

}

}

//main
public class Benchmarking {


public static void main (String args[]){

Scanner scanner = new Scanner(System.in);

     System.out.println("What is the capacity of your Array?");
     int capacity = scanner.nextInt();

   methods array1 = new methods(capacity);
   System.out.println(array1.getArray());
 }

运行程序并输入容量后,它崩溃了。我只需要创建一个arrayList,用随机数填充它并打印出来。以下是我收到的错误列表:

Exception in thread "main" java.util.IllegalFormatConversionException:  d   != java.util.ArrayList
at java.util.Formatter$FormatSpecifier.failConversion(Formatter.java:4302)
at java.util.Formatter$FormatSpecifier.printInteger(Formatter.java:2793)
at java.util.Formatter$FormatSpecifier.print(Formatter.java:2747)
at java.util.Formatter.format(Formatter.java:2520)
at java.util.Formatter.format(Formatter.java:2455)
at java.lang.String.format(String.java:2927)
at Benchmarking.methods.getArray(methods.java:68)
at Benchmarking.Benchmarking.main(Benchmarking.java:27)

我认为我的方法做了一些根本性的错误。

2 个答案:

答案 0 :(得分:0)

this.capacity = capacity;添加到public methods() {构造函数中以开始。您正在引用此变量但从未设置它。

答案 1 :(得分:0)

您无法将randomArrayjava.util.ArrayList}传递给String.format()

您可能希望改为通过randomArray.get(i)