How to print ASCII value of an int in JAVA

Question

I've searched for this on the internet but was unable to find a precise solution, one possible solution I found was to read the integer as a String, and use charAt method and then cast the character to int to print the ascii value.

But is there any other possible way to do it other than the above stated method?

int a=sc.nextInt();
//code to convert it into its equivalent ASCII value.

For example, consider the read integer as 1, and now I want the ASCII value of 1 to be printed on the screen,which is 49

Why don't you show a working example so we have a clearer idea of what you're trying to do? — shmosel, Sep 18 '16 at 09:55
Java characters are Unicode which covers many more characters than ASCII — greg-449, Sep 18 '16 at 09:59
Possible duplicate of [Convert string to ASCII numeric value in java](http://stackoverflow.com/questions/16458564/convert-string-to-ascii-numeric-value-in-java) — Michał Szewczyk, Sep 18 '16 at 12:29

score 3 · Answer 1 · answered Sep 18 '16 at 09:53

3

I assume you're looking for this:

System.out.print((char)a);

answered Sep 18 '16 at 09:53

shmosel

49,289
6
73
138

score 1 · Answer 2 · answered Dec 31 '17 at 15:49

The easy way to do that is:

For the Whole String

public class ConvertToAscii{
    public static void main(String args[]){
      String number = "1234";
      int []arr = new int[number.length()];
      System.out.println("THe asscii value of each character is: ");
      for(int i=0;i<arr.length;i++){
          arr[i] = number.charAt(i); // assign the integer value of character i.e ascii
          System.out.print(" "+arr[i]);
      }
    }
}

For the single Character:


String number="123";
asciiValue = (int) number.charAt(0)//it coverts the character at 0 position to ascii value

How to print ASCII value of an int in JAVA

2 Answers2

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