CODEWITHSUNDEEP
javaarrays
user7590445
user7590445

Reputation:

Why are there extra values in an array?

My code is here:

import java.util.Arrays;

public class Q1102 {
    public static void main(String[] args) {
        String temp = "1 3 5 7 6";
        int[] array1 = new int[temp.length()];
        for (int i = 0; i < temp.length(); i += 2) {
            int mounth = Integer.parseInt(String.valueOf(temp.charAt(i)));
            array1[i] = mounth;
        }
        System.out.println(Arrays.toString(array1));
    }
}

For some strange reason, the program outputs this:

[1, 0, 3, 0, 5, 0, 7, 0, 6]

Can someone please tell me why this occurs, and how to remove the extra zeros?

I've printed out the values that entered the array as they are entered, and there aren't any extra zeros.

Upvotes: 0

Views: 124

Answers (2)

chandan gupta
chandan gupta

Reputation: 1349

Here issue in this line temp.length(). It will return 9 as a string length. As you know java takes default value 0. That's why you are getting extra zeros.

import java.util.Arrays;

public class Q1102 {
public static void main(String[] args) {
    String temp = "1 3 5 7 6";
    int length = (temp.length()  + 1)/2;
    int[] array1 = new int[length];
    for (int i = 0, k = 0; i < temp.length(); i +=2, k++) {
        int mounth = Integer.parseInt(String.valueOf(temp.charAt(i)));
        array1[k] = mounth;
    }
    System.out.println(Arrays.toString(array1));

}

Upvotes: 0

Elliott Frisch
Elliott Frisch

Reputation: 201439

You are using character conversion in a way that only works with single digit numbers, it's fragile in a number of ways. Instead, split the String on whitespace and then parse those tokens. Like,

String temp = "1 3 5 7 6";
String[] tokens = temp.split("\\s+");
int[] array1 = new int[tokens.length];
for (int i = 0; i < tokens.length; i++) {
    array1[i] = Integer.parseInt(tokens[i]);
}
System.out.println(Arrays.toString(array1));

Outputs (as I think you expected)

[1, 3, 5, 7, 6]

Upvotes: 4

Related Questions