Char in a switch - java 7

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1.00/5 (1 vote)

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I have the below code:

Where do i wrong?

I am confused because i is char.

What I have tried:

Java

public class SwitchChar {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        char i;
        LOOP:
        for (i = 0; i < 5; i++) {
            System.out.println((int)i);
            switch (i++) {
                case '0':
                    System.out.println((int)i);
                    System.out.println("A");
                case 1:
                    System.out.println((int)i);
                    System.out.println("B");
                    break LOOP;
                case 2:
                    System.out.println((int)i);
                    System.out.println("C");
                    break;
                case 3:
                    System.out.println((int)i);
                    System.out.println("D");
                    break;
                case 4:
                    System.out.println((int)i);
                    System.out.println("E");
                case 'E':
                    System.out.println((int)i);
                    System.out.println("F");
            }
        }
    }

}

I don't understand the steps who conduct to the result.

when i = 0 - no case to select, i is incremented to 1
when i = 1 - break loop so for is never executed

Thank a lot!!!

Posted 20-Feb-17 7:25am

Anda Cristea

Updated 20-Feb-17 11:38am

Patrice T

v2

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2 solutions

Solution 2

You do understand the common fact that zero is not same in integer form, and character form. Their binary, that gets generated and checked up against is not same. A character is a switch-able value, but it will not map any true value for the cases that you are using it for.

To solve this problem, change one of the types to other one,

for (i = 0; i < 5; i++) {
    System.out.println((int)i);
        switch (i++) {
            case 0:
            // Code here.
            // ...

Otherwise, you should consider using character values, and use Character.valueOf(i); and then move onwards. That would look something like this,

for (i = 0; i < 5; i++) {
     // Map it.
     char ch = Character.valueOf(i);
     System.out.println(i); // No need of (int)int; it is already an int.
         switch (ch++) {
             case '0':
             // code here...

Doing the ch++ is legal in Java, because a character is a numeric value that represents the Unicode character. Incrementing it, provides you with the next character and so on. If that doesn't sound easy, read about pointer arithmetic.

Posted 20-Feb-17 11:38am

Afzaal Ahmad Zeeshan

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Patrice T · Accepted Answer · 2017-02-20T09:25:00

When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]
http://docs.oracle.com/javase/7/docs/technotes/tools/windows/jdb.html[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[^]

The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.