Why is my charAt when printed giving me a number instead of a letter

Question

When using the code below, I am a beginner testing what String methods do. Though, whenever using hello and using a char i to represent 0, why is it not printing out h and instead 104. And if there is a reason why it is printing 104, why?

import java.util.Scanner;

public class Test {
    public static void main(String[] args) {
        String hello = "hello";
        char i  = hello.charAt(0);
        System.out.println(+i);
    }
}

To expand on that:correct comment (which should IMO be written up as the answer) `+i` widens the `char` to an `int`, which is formatted differently. — user15187356, Mar 12 '21 at 04:48
Specifically, I think the unary operator `+` does what most other math does in Java: it returns an integer, not a `char`. You'd have to convert the number back: `System.out.println( (char)(+i) );` — markspace, Mar 12 '21 at 04:49
This seems relevant: https://stackoverflow.com/questions/2624410/what-is-the-purpose-of-javas-unary-plus-operator#2624541 — xdhmoore, Mar 12 '21 at 04:52

score 3 · Answer 1 · answered Mar 12 '21 at 04:52

The '+' is messing you up. Remove that one character and you'll get the result you desire:

class Solution {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String hello = "hello";
        char i  = hello.charAt(0);
        System.out.println(i);
    }
}

Result:

104 is the ASCII value that represents the 'h' character. I couldn't have told you this in advance, but apparently putting a'+' in front of a character value causes it to be converted to an integer that is its ASCII value.

Why is my charAt when printed giving me a number instead of a letter

1 Answers1