I cast the scanners int to a char why is this giving me a mismatch exception?

Question

public class Practice {
public static void main(String []args){
Scanner ScanMe=new Scanner(System.in);
char ch,answer='K';
System.out.println("I am thinking of a letter between A and Z.");
ch=(char)ScanMe.nextInt();
System.out.println(ch);
if(answer==ch){
System.out.println("CORRECT");
}
}
}

I created a new scanner, I created my char variables and then I read my char variables in and its giving a mismatch error???

Possible Duplicate : http://stackoverflow.com/questions/13942701/take-a-char-input-from-the-scanner — Jitin Kodian, Feb 10 '16 at 04:37

score 2 · Answer 1 · answered Feb 10 '16 at 04:58

2

Matt's solution is correct, but I wanted to add a bit of clarification. ch=(char)ScanMe.nextInt(); throws an exception because ScanMe.nextInt() gets evaluated first. Assuming you entered a letter, nextInt would throw an exception because a letter is not a decimal integer (which nextInt expects).

However, things will work if you enter a decimal integer. For instance, if you enter 75, this method call will work, and when you cast 75 to a char, it's actually 'K', so your program as originally written would say "CORRECT".

Check an ASCII table if that casting doesn't make sense.

answered Feb 10 '16 at 04:58

tyler

91
5

Oh I got it, so basically once the nextInt realizes I put anything other then an int in it automatically throws an exception and does not get cast to a char? – Eli Feb 10 '16 at 05:01
1

That's pretty much it. As soon as the exception is thrown, no more of the code is executed (until the exception is caught, if ever). – tyler Feb 10 '16 at 05:07

score 1 · Accepted Answer · answered Feb 10 '16 at 04:42

1

Try using this to get a char

ch = ScanMe.next().charAt(0);

also K is not the same as k.

answered Feb 10 '16 at 04:42

Matt

52
7

Dude great answer, I will give u the check when it allows, also why did you put zero in the .charAt, what does that accomplish – Eli Feb 10 '16 at 04:44
charAt(0) makes sure it only takes the first charter you enter so if you enter abscdsfdfhfgj it only takes the a – Matt Feb 10 '16 at 04:46

Scary Wombat · Answer 3 · 2016-02-10T04:44:28.773

0

An Integer is not a char

Try using next and then just using the first char of the String

use String.charAt(int);

https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#charAt(int)

edited Feb 10 '16 at 04:44

answered Feb 10 '16 at 04:39

Scary Wombat

41,782
5
32
62

but I cast it to an char – Eli Feb 10 '16 at 04:40
Thats what I said, also you cannot cast from a string to a char, the first letter of a string is not the same as a char value – Eli Feb 10 '16 at 04:41
use `charAt` https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#charAt(int) – Scary Wombat Feb 10 '16 at 04:42
how do cast 12345 to a char? – Scary Wombat Feb 10 '16 at 04:43

I cast the scanners int to a char why is this giving me a mismatch exception?

3 Answers3