Something wrong in "Character Count Java Program"

Question

This program is supposed to count the number of characters entered by a user. Where other is other characters such as !, @, $, etc. It is not supposed to count #. The following is my code to do this:

public class countchars {
    public static void main(String args[]) {
        Scanner input = new Scanner(System.in);

        char sym;
        int up = 0;
        int low = 0;
        int digit = 0;
        int other = 0;

        System.out.print("Enter a character # to quit: ");
        sym = input.next().charAt(0);

        while (sym != '#') {
            System.out.print("Enter a character # to quit: ");

            if (sym >= 'a' && sym <= 'z') {
                low++;
            }
            if (sym >= 'A' && sym <= 'Z') {
                up++;
            }
            if (sym >= '0' && sym <= '9') {
                digit++;
            }
            if (sym >= '!' && sym <= '=') {
                other++;
            }

            sym = input.next().charAt(0);

        }

        System.out.printf("Number of lowercase letters: %d\n", low);
        System.out.printf("Number of uppercase letters: %d\n", up);
        System.out.printf("Number of digits: %d\n", digit);
        System.out.printf("Number of other characters: %d\n", other);
    }
}

The problem is with the "other" counter. If I entered !, @, and $, it will only count 2 of the 3 characters entered. What's the wrong?

yurib · Accepted Answer · 2013-03-07 16:00:03Z

3

if you take a look at the ascii table, you'll see that:
'!' = 33
'=' = 61
'@' = 64

the '@' character isn't in the range you specified so it's not counted, replace the last condition with:

if (sym >= '!' && sym <= '@') {...}

answered Mar 7, 2013 at 16:00

yurib

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Comments

CsBalazsHungary · Accepted Answer · 2013-03-07 16:07:29Z

1

try this:

        if (sym >= 'a' && sym <= 'z') {
            low++;
        } else if (sym >= 'A' && sym <= 'Z') {
            up++;
        } else if (sym >= '0' && sym <= '9') {
            digit++;
        } else {
            other++;
        }

or instead of else you can select the short set of what that character can be:

        } else if ("%!$&".contains(sym)){
            other++;
        }

answered Mar 7, 2013 at 16:07

CsBalazsHungary

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1 Comment

CsBalazsHungary Over a year ago

to add an explanation: the special symbols may be excluded from the ascii range of "!" and "="

m0skit0 · Accepted Answer · 2013-03-07 16:43:10Z

1

Try with

else {
    other++;
}

instead of

if (sym >= '!' && sym <= '=') {
    other++;
}

# will not be counted as other because you already filter it in the while condition.

edited Mar 7, 2013 at 16:43

answered Mar 7, 2013 at 15:58

m0skit0

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Comments

Subhrajyoti Majumder · Accepted Answer · 2013-03-07 15:57:44Z

0

You should use OR(||) in condition instead of AND(&&)

if (sym == '!' || sym == '=' || sym == '@' || ...){
        other++;
}

answered Mar 7, 2013 at 15:57

community wiki

Subhrajyoti Majumder

Comments

Henry Keiter · Accepted Answer · 2013-03-07 16:01:02Z

0

To be sure you catch "everything else"; you just use an else clause. That way you don't miss things, like you're doing now (because '@' isn't in the range you're checking). You want this:

else {
    other++;
}

where you currently have this:

if (sym >= '!' && sym <= '=') {
    other++;
}

answered Mar 7, 2013 at 16:01

Henry Keiter

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Comments

BobTheBuilder · Accepted Answer · 2013-03-07 16:01:22Z

0

You compare char based on the ASCII value. @ ASCII is 64 ! ASCII is 33 = ASCII is 61

So @ isn't between "!" and "=" and doesn't increment your counter.

answered Mar 7, 2013 at 16:01

BobTheBuilder

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Comments

PelleLauritsen · Accepted Answer · 2013-03-07 16:03:03Z

0

Look here, and hopefully the answer will present it self!

http://en.wikipedia.org/wiki/UTF-8

and are you sure the sequence $,5,$ will give you the right answer ? ;)

answered Mar 7, 2013 at 16:03

PelleLauritsen

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