CODEWITHSUNDEEP
javaarraysstringint
shanahobo86
shanahobo86

Reputation: 497

Java string to Int

I have a string with about 150 numbers like this String num = "64513246563........";

I am trying to add each digit of this string. So my idea was to split it into an array of ints and add them from there. I start by splitting it into a String array, then I try to convert it to an Int array. I get an unknown source error. Below is the code:

String[] strArray = num.split("");
int[] intArray = new int[strArray.length];
        for(int i = 0; i < strArray.length; i++) {
           intArray[i] = Integer.parseInt(strArray[i]);
           }

And here is the error:

Exception in thread "main" java.lang.NumberFormatException: For input string: ""
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)

Can anyone see what I'm doing wrong or is there a more efficient way of doing this?

////////////////////////////////

Thanks for your help everyone, it seems that splitting a string using .split("") creates an empty string at index 0. That was my main problem but there was lots of useful pointers on how to solve the problem more efficiently :) Thank you all for your input

Upvotes: 3

Views: 7022

Answers (8)

user219882
user219882

Reputation: 15844

int sum = 0;
for (int i = 0; i < string.length(); i++) {
    sum += Character.getNumericValue(string.charAt(i));
}

Upvotes: 5

someone
someone

Reputation: 6572

Since array start from a empty string you have to skip first element hence Start i from 1 in your for loop.

for(int i = 1; i < strArray.length; i++)

Upvotes: 0

PermGenError
PermGenError

Reputation: 46408

String num="12234";
        String[] numarr = num.split("");
        System.out.println(numarr.length);

When you split a string with "" there will be an empty string at the zeroth index. Thus your resulting array would be :

{"", "1", "2", "2", "3", "4"}

test it, the above returns the length as 6. thus Integer.parseInt("") leads to NumberFormatException.

Test:

String num="12234";
        String[] numarr = num.split("");
        for(String s: numarr){
            if(s.equals("")){
                System.out.println("empty string");
            }
            else {
                System.out.println(s);
            }

Output:

empty string
1
2
2
3
4

Thus, there is an emptyString at index 0.

so, you will have to check if the String is an empty string before parsing it:

String[] strArray = num.split("");
int[] intArray = new int[strArray.length];
        for(int i = 0; i < strArray.length; i++) {
           if(!strArray[i].equals(""))
           intArray[i] = Integer.parseInt(strArray[i]);
           }

Upvotes: 2

kosa
kosa

Reputation: 66637

java.lang.NumberFormatException: For input string: ""

exception message clearly telling that input is empty String.

One way to handle would be, do empty String check before parsing.

 for(int i = 0; i < strArray.length(); i++) {
           if(strArray[i] != null && !"".equals(strArray[i])
              {
               int q = Integer.parseInt(strArray[i]);
              }
           }

Note: This will still fail if input is not valid int, but not in case of empty String (or) null.

Another way to handle this would be get chartAt(i) instead of using split("")

Upvotes: 3

Ted Hopp
Ted Hopp

Reputation: 234795

I'd simply step along the string, parsing each digit:

int sum = 0;
for (int i = 0; i < num.length(); ++i) {
    sum += Integer.parseInt(num.substring(i, i + 1));
}

Upvotes: 0

cowls
cowls

Reputation: 24334

char[] charArray = num.toCharArray();

for(char c : charArray) {
    int q = Integer.parseInt("" + c);
}

Upvotes: 1

Denys S&#233;guret
Denys S&#233;guret

Reputation: 382150

The first string given by num.split(""); is empty.

A solution is to simply start the iteration at index 1.

But what you do is much too expensive as you could do

    String num = "64513246563";
    for(int i = 0; i < num.length(); i++) {
         int digit = num.charAt(i)-'0';
         ... // use the digit, perhaps adding it
    }

Upvotes: 8

İsmet Alkan
İsmet Alkan

Reputation: 5447

for(int i=0; i<yourString.length; i++) {
    yourIntegers[i] = Integer.parseInt(yourString.charAt(i).ToString());
}

Upvotes: 0

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