java无法捕获从一个方法到另一个方法的返回值
我目前正在用Java创建一个小应用程序。然而,我现在被卡住了。我很难从一个方法到另一个方法捕获返回值
以下是我何时返回值以及在何处获取值的示例:
public int[] highest(int y[]){
int largest = 500;
int largeArr[] = new int[y.length];
for(int count = 0; count < y.length; count++)
{
if(y[count] >= largest){
largeArr[count] = y[count];
System.out.print(largeArr[count] + " ");
}
}
return largeArr;
}
public void highText()
{
RandomNumbers high = new RandomNumbers();
System.out.println("\nDessa tal är " + high.highest() + " intervallet 500 - 999: ");
}
我很清楚,我应该利用参数和参数。但是,当我以后想要打印文本时,这就成了一个问题
完整代码:
RandomNumbers rnd = new RandomNumbers();
rnd.highText();
rnd.highest(nummer);
import java.util.Scanner;
public class RandomNumbers {
public void rnd(int y[]){
for(int count = 0; count < y.length; count++)
{
y[count] = (int) ( Math.random()*1000);
System.out.print(y[count] + " ");
}
}
public int[] lowest(int y[]){
int largest = 499;
int smallArr[] = new int[y.length];
for(int count = 0; count < y.length; count++)
{
if(y[count] <= largest){
smallArr[count] = y[count];
System.out.print(smallArr[count] + " ");
}
}
return smallArr;
}
public int[] highest(int y[]){
int largest = 500;
int largeArr[] = new int[y.length];
for(int count = 0; count < y.length; count++)
{
if(y[count] >= largest){
largeArr[count] = y[count];
System.out.print(largeArr[count] + " ");
}
}
return largeArr;
}
public void highText()
{
RandomNumbers high = new RandomNumbers();
System.out.println("\nDessa tal är " + high.highest() + " intervallet 500 - 999: ");
}
public void lowText()
{
System.out.println("\nDessa " + smallArr + " tal är i intervallet 0 - 499: ");
}
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
System.out.print("Hur många slumptal i intervallet 0-999 önskas? ");
int x = user_input.nextInt();
int nummer[] = new int[x];
RandomNumbers rnd = new RandomNumbers();
System.out.println("\nHär är de slumpade talen: ");
rnd.rnd(nummer);
rnd.lowText();
rnd.lowest(nummer);
rnd.highText();
rnd.highest(nummer);
}
}
当我打印文本时,我使用rnd方法。highText();。如果我之前在参数中使用Largear,那么程序会请求参数——当我真的只想打印方法时:(
请帮忙()
共 (0) 个答案