Arithmetic promotion

[dhruv simaria]

Simon Roberts in his book Java 2 certification guide mentions with reference to Unary operators that
There are 2 rules that apply depending on the type of the operand
1)If one of the operand is a byte, a short or a char,it is converted to an int
using this rule the code given below when compiled should give a type mismatch error as converting an int to a char requires a cast.Surprisingly not only does this code compile but when it is run,The output is - The character is now b
Why does this hapen ?

class Achtung {
public  static void  main(String args[]) {
char A='a';
++A;
System.out.println("The  character  is  now "+A);  }
                }

[This message has been edited by dhruv simaria (edited March 27, 2001).]

[Marilyn de Queiroz]

These operators use the native CPU increment or decrement which is unsigned. You will note a similar operation using byte does not require a cast.

However, this topic is unrelated to the assignments here in the cattle drive so I am moving it to Java In General (Beginning)

[Richard Boren]

class CharTest
{
   static public void main( String args[] )
   {
        int x = 97 ;      // decimal value of 'a' in ASCII table
        char y = (char)x ;  // cast required since char is                             // unsigned 16-bit (smaller)
                            // and int is signed 32-bit (larger).                              // The result is (smaller).
        System.out.println( "Here is y: " + y ) ;  // prints 'a'
        // Now prints 97
        System.out.println( "Here is y int val: " + (int)y ) ;          

        y++;
        // prints 'b'
        System.out.println( "Here is y after ++:  " + y ) ; 
        // print 98
        System.out.println( "Here is y int val:  " + (int)y ) ;  
   }
}

I don't think char is actual converted to a int during any arithmetic operation.
According to Peter van der Linden in Just Java 2 page 96,
char is an integer-based type with all arithmetic operators available. The code above will first produce an 'a' since x is assigned 97 (the decimal value of 'a' in the ASCII table).
The char casting for x is required since char is a unsigned 16-bit type and int is a signed 32-bit type so the result of the assignment will be smaller. If y is then cast to int it will print 97. Any arithmetic done to y is performed as if y were an integer(char is a integer-base type).
Hope this is right, makes sense and helps.