将 Object[] 数组转换为 java 中的 int [] 数组？

Question

It appears there is no easy way of doing this, but this is what i've done so far and if someone could correct it to make it work that would be great. At "newarray [e] = array [i].intValue ();"i get an error "No method named "intValue" was found in type "java.lang.Object"." Help!

/*
Description: A game that displays digits 0-9 and asks the user for a number N.
 It then reverses the first N numbers of the sequence. It continues this until
 all of the numbers are in order.
 numbers

*/

import hsa.Console;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Arrays;


public class ReversalGame3test

{
    static Console c;

    public static void main (String[] args)
{
    c = new Console ();

    c.println ("3. REVERSAL GAME");
    c.println ("");
    c.println ("Displayed below are the digits 0-9 in random order. You must then enter a");
    c.println ("number N after which the computer will reverse the first N numbers in the");
    c.println ("sequence. The goal of this game is to sort all of the numbers in the fewest");
    c.println ("number of reversals.");
    c.println (""); //introduction

    List numbers = new ArrayList ();
    numbers.add ("0");
    numbers.add ("1");
    numbers.add ("2");
    numbers.add ("3");
    numbers.add ("4");
    numbers.add ("5");
    numbers.add ("6");
    numbers.add ("7");
    numbers.add ("8");
    numbers.add ("9");
    Collections.shuffle (numbers);
    Object[] array = numbers.toArray (new String [10]); // declares + shuffles numbers and converts them to array

    c.print ("Random Order: ");
    for (int i = 0 ; i < 10 ; i++)
    {
        c.print ((array [i]) + " ");
    }
    c.println ("");

    boolean check = false;
    boolean check2 = false;
    String NS;
    int N = 0;
    int count = 0;
    int e = -1;
    int[] newarray = new int [10];

    //INPUT
    do
    {
        c.print ("Enter a number: ");
        NS = c.readString ();
        count += 1;

        check = isInteger (NS);
        if (check == true)
        {
            N = Integer.parseInt (NS);
            if (N < 1 || N > 10)
            {
                check = false;
                c.println ("ERROR - INPUT NOT VALID");
                c.println ("");
            }
            else
            {
                c.print ("Next Order: ");
                for (int i = N - 1 ; i > -1 ; i--)
                {
                    e += 1;
                    newarray [e] = array [i].intValue ();
                    c.print ((newarray [e]) + " ");
                }
                for (int i = N ; i < 10 ; i++)
                {
                    e += 1;
                    newarray [e] = array [i].intValue ();
                    c.print ((newarray [e]) + " ");
                }
                check2 = sorted (newarray);
            } // rearranges numbers if valid
        } // checks if N is valid number
    }
    while (check == false);
} // main method


public static boolean isInteger (String input)
{
    try
    {
        Integer.parseInt (input);
        return true;
    }
    catch (NumberFormatException nfe)
    {
        return false;
    }
} //isInteger method


public static boolean sorted (int array[])
{
    boolean isSorted = false;

    for (int i = 0 ; i < 10 ; i++)
    {
        if (array [i] < array [i + 1])
        {
            isSorted = true;
        }
        else if (array [i] > array [i + 1])
        {
            isSorted = true;
        }
        else
            isSorted = false;

        if (isSorted != true)
            return isSorted;
    }
    return isSorted;
} // sorted method

}

Answer 1

You can use Integer.valueOf .

Integer.valueOf((String) array [i])

The Integer class has a method valueOf which takes a string as the value and returns a int value, you can use this. It will throw an NumberFormatException if the string passed to it is not a valid integer value.

Also If you are using java5 or higher you can try using generics to make the code more readable.

Answer 2

You can implement the same using Generics , which would be easier.

List<Integer> numbers = new ArrayList<Integer> ();
Integer[] array = numbers.toArray (new Integer [10]);

Answer 3

试试 commons-lang

org.apache.commons.lang.ArrayUtils.toPrimitive(Integer[])

Answer 4

You can't call .intValue() on an Object , as the Object class lacks the method intValue() .

Instead, you need to cast the Object to the Integer class first, like so:

newarray[e] = ((Integer)array[i]).intValue();

Edit: Just a helpful tip on StackOverflow - please limit your code to what's relevant! Though sometimes large blocks of code are necessary, in this case, it was not. It makes the question look nicer, and it's bound to get better responses that way.

Also, please do not use the homework tag. It is currently deprecated and is in the process of burnination.

Answer 5

I know this is pretty late, but here are my 2 cents!!

int[] newArray=new int[objectArray.length];
Object[] objectArray = {1,2,3,4,5};

for(int i=0;i<objectArray.length();i++){
    b[i]=(int)objectArray[i];
}

Answer 6

I made this method... I think is better!

public int[] ToMixArray(Object[] Array, int StratIndex, int Valuedefault, int NewLength){

    int[] res=new int[NewLength];
    for (int i = 0; i < NewLength; i++) {
        try { res[i]=Integer.parseInt(String.valueOf(Array[StratIndex+i]));}
        catch(Exception e){res[i]=Valuedefault;}
    }return res;
}