CODEWITHSUNDEEP
java
Sean Roh
Sean Roh

Reputation: 1

Why does this keep printing out 51?

public class Main
{
    public static void main(String[] args) 
    {
        System.out.println(digitofPi(0));
    }
    public static int digitofPi(int digit)
    {
        String x = String.valueOf(Math.PI);
        char result = x.charAt(digit);
        return result;
    }
}

I'm trying to print out nth position of pi by converting pi to string, however it keeps printing out 51.

Upvotes: 0

Views: 190

Answers (4)

zslevi
zslevi

Reputation: 409

You're converting the character to int, which means you're getting it's ASCII code. (Actually Unicode codepoint, but for your case it's the same.)

http://www.asciitable.com/

If you look at it, you'll see that the code of "0" is 48 and the code of "3" is 53. So you'd have to subtract 48 from it.

Upvotes: 0

Looken
Looken

Reputation: 75

This one will return 51

public static int digitofPi(int digit) {
    String x = String.valueOf(Math.PI);
    return x.charAt(digit);
}

Upvotes: -2

Frakcool
Frakcool

Reputation: 11153

Let's take your code step by step and let's consider that Pi = 3.1415

Math.PI = 3.1415

So, when you're

String x = String.valueOf(Math.PI);

Then

x = "3.1415"

And when you do this:

char result = x.charAt(digit);

Then result becomes '3'

And if you return an int from a char, it's going to take its ASCII value, so:

'3' = 51

So you can either return char or String on your method

return result

Upvotes: 3

Unmitigated
Unmitigated

Reputation: 89294

You want to return a char, not an int, unless you want to display the ASCII character code of the digit.

public static char digitofPi(int digit) {
    String x = String.valueOf(Math.PI);
    char result = x.charAt(digit);
    return result;
}

Note that you should remove the decimal point from the String if you do not want it to be treated as the second digit.

String x = String.valueOf(Math.PI).replace(".", "")

Upvotes: 1

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