String array with letters to int convertion

Question

I got this array of string

String s1[] = {"H","E","L","L","O"};

and a method to convert it into int

    public static int[] String2Int(String[] k){
    int[] arrayBuffer = new int[k.length];

    for (int i=0; i < k.length; i++)
    {
        //arrayBuffer[i] = Integer.parseInt(k[i]);
        arrayBuffer[i] = Integer.parseInt(k[i]);
    }
    return arrayBuffer;

I want the decimal values of the characters but no luck on getting it. Google is only helping with if the string were a number.

What do you mean decimal values of the characters? you mean ascii value of character or parsing like "43" = 43? or 'A' = 65 ? — Abdullah Tellioglu, Mar 19 '18 at 08:09
https://stackoverflow.com/questions/2006533/how-can-i-get-a-unicode-characters-code#2006580 — Tschallacka, Mar 19 '18 at 08:10
Possible duplicate of [How can I get a Unicode character's code?](https://stackoverflow.com/questions/2006533/how-can-i-get-a-unicode-characters-code) — ernest_k, Mar 19 '18 at 09:52

Suresh Atta · Accepted Answer · 2018-03-19T08:21:32.313

2

You shouldn't convert it to integer. It should be

 arrayBuffer[i] =k[i].charAt(0);

That way you get the ASCII value of char and gets assigned to int.

Edit :

You can also use arrayBuffer[i] = Character.codePointAt(input, 0); as pointed in comments.

edited Mar 19 '18 at 08:21

answered Mar 19 '18 at 08:10

Suresh Atta

120,458
37
198
307

why not `Character.codePointAt(input, 0);` – Tschallacka Mar 19 '18 at 08:11
@Tschallacka well both works. Isn't it? – Suresh Atta Mar 19 '18 at 08:14
It handles also the edge cases where it isn't always straightforward what code point it is, it handles surrogate pairs so you'll get a consistent output back. – Tschallacka Mar 19 '18 at 08:18
arrayBuffer[i] =k[i].charAt(0); Did the trick for ruinning the program. thanks! – Tor Mar 19 '18 at 08:20
1

@Tschallacka Fair enough. Updated my answer. – Suresh Atta Mar 19 '18 at 08:22

Impulse The Fox · Answer 2 · 2018-03-19T08:25:33.133

Try this:

class MainClass
{
    public static void main(String[] args)
    {
        System.out.println(Arrays.toString(String2Int("HELLO")));
    }
    public static int[] String2Int(String k)
    {
        int[] arrayBuffer = new int[k.length()];

        for (int i = 0; i < arrayBuffer.length; i++)
        {
            //arrayBuffer[i] = Integer.parseInt(k[i]);
            arrayBuffer[i] = k.charAt(i);
        }
        return arrayBuffer;
    }
}

Output:

[72, 69, 76, 76, 79]

Explanation

char and int is almost the same in Java. You don't need to parse it, you just implicitly cast it by asigning it to the int[]. Also you shouldn't take a String[] as an argument, that makes no sense. A String is already almost a char[]. So either take a String, or a char[]. In my example I've used a String because that is more convenient to call.

score 0 · Answer 3 · answered Mar 19 '18 at 09:14

Using the Java 8 Stream API:

public static int[] convert(String[] strings)
{
    Objects.requireNonNull(strings);
    return Arrays.stream(strings)
            .mapToInt(str -> Character.getNumericValue(str.charAt(0)))
            .toArray();
}

I assume you dont need to check that the input strings contain exactly one character.

String array with letters to int convertion

3 Answers3

Try this:

Output:

Explanation