Displaying reverse numbers array

Question

So I have to write a program that accepts 10 numbers (ints) from the keyboard. Each number is to be stored in a different element of an array. Then my program must then display the contents of the array in reverse order.

int [] array = new int [10];
    for(int i = array.length -  1;i >= 0; i--)
    {
    int number = Integer.parseInt(JOptionPane.showInputDialog(null, "Enter number " + (i+1)));
        array[i] = number;

            }
                JOptionPane.showMessageDialog(null, array[i]);


}

I tried to put the JOPtionPane.showMessageDialog outside of the loop but then the program can't find the integer "i". I don't know what to do here :/ Please help :P

You need two for loops. The first iterates from 0 to 9 and asks for the number and puts it in the array. The second iterates from 9 to 0 and prints the numbers in the array. — Mark
– Mark, Commented Mar 17, 2016 at 20:06

ManoDestra · Accepted Answer · 2016-03-17 20:13:18Z

2

You need to enter your data first, then display it thereafter in the order you desire...

int [] array = new int[10];
for (int i = 0; i < array.length - 1; i++) {
    int number = Integer.parseInt(JOptionPane.showInputDialog(null, "Enter number " + (i + 1)));
    array[i] = number;
}

for (int i = array.length - 1; i >= 0; i--) {
    JOptionPane.showMessageDialog(null, array[i]);
}

I'd also be tempted to simply construct a StringBuilder for your final results and then just show the message dialog once only, rather than for every element of the array, but that's up to yourself :)

edited Mar 17, 2016 at 20:13

answered Mar 17, 2016 at 20:09

ManoDestra

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Comments

Guy · Accepted Answer · 2016-03-17 20:10:19Z

1

i belongs to the loop scope, that's why you can't use it outside of the loop.

To print the reversed array use another loop

// insert the data to the array
int [] array = new int [10];
for(int i = array.length -  1 ; i >= 0 ; i--) {
    int number = Integer.parseInt(JOptionPane.showInputDialog(null, "Enter number " + (i+1)));
    array[i] = number;
}

// print the array
for (int i = 0 ; i < array.length ; ++i) {
    JOptionPane.showMessageDialog(null, array[i]);
}

answered Mar 17, 2016 at 20:10

Guy

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3 Comments

hDDen Over a year ago

Thanks for this, I have another question, I input numbers 10,9,8,7,6,5,4,3,2 (which is 9th number) and then it transfers me to another loop where I get output of 0,2,3,4,5,6,7,8,9,10. Why does it cut 1 number? You have any idea?

Guy Over a year ago

@PatrykMichta I'm not sure, I ran it in my computer and it was fine.

hDDen Over a year ago

I got it, it was my bad. Thanks anyway :)

Z-DNA · Accepted Answer · 2016-03-17 20:11:09Z

0

That's because you've declared in in for loop, so it has loop scope. Declare it before loop to reuse it after loop finishes

int i;
for(i = array.length -  1;i >= 0; i--)

After that, you can make another loop:

for(i = 0; i < array.length; i++)

to print it in reverse order.

answered Mar 17, 2016 at 20:11

Z-DNA

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2 Comments

hDDen Over a year ago

Thanks for this, I have another question, I input numbers 10,9,8,7,6,5,4,3,2 (which is 9th number) and then it transfers me to another loop where I get output of 0,2,3,4,5,6,7,8,9,10. Why does it cut 1 number? You have any idea?

Z-DNA Over a year ago

Oh, that's because second for prints whole table (which size was 10, so it print from index 0 to index 10) but you only have filled from index 9 to index 1, so the unfilled fields in array was 0 (initialization value)

Keision Palms · Accepted Answer · 2016-03-17 20:11:25Z

0

You need two for loops. The first iterates from 0 to 9 and asks for the number and puts it in the array. The second iterates from 9 to 0 and prints the numbers in the array

answered Mar 17, 2016 at 20:11

Keision Palms

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