public	class Class1
{			
	public static void main(String args[])
	{		
		int i =1; 
		char c = '2';
		c += i;                // i += c;    
		System.out.println("char: " + c); // System.out.println("int: " + i);
    }
}

o/p:
char: 3
int: 51 //if I use the commented line in place of the original

In the second case ie;
i += c;
Integer value of char '2' is 50 (from ASCII table)
(Also, I know chars' in java are Unicode)

So, the above stmt evaluates the value of 'i' to 51
and then prints it.

Is there anything more to this ?

Regds.

Check these out:

byte b1=1, b2 = 2;
byte b = b1 + b2; // Compile Error.
// int cannot be assigned to byte.

int i = 1;
byte b3 = b1 + i; // Compiles Okay and result is 2

Same if you replace byte with short.

Also,
b1 += 1; /Compiles and result is 2.

Not sure whats going on here .....

Also note that in order for the String conversion of the other operand to take place, one of the 2 operands MUST be a String object. In other words, aStringBufferObject + primitive WILL NOT work.

Simillarly there is a corresponding String conversion for the literals true/false/null. But they can be either used individually / (literal +String) for the System.out.println(..) method in order to work. In other words,
System.out.println(true); //prints 'true'
System.out.println(false); //prints 'false'
System.out.println(null); //prints null
System.out.println("maha"+null); //prints 'mahanull'
System.out.println("Umesh"+true); //prints 'Umeshtrue'

System.out.println(false+true); //NOT OK
System.out.println(false+null); //NOT OK
System.out.println(true+null); //NOT OK

StringBuffer sb= new StringBuffer("java");
System.out.println(sb+10); //NOT OK because atleast 1 must be String

So, coming back to your qstn,
System.out.println("char: " + c);
In the above statement for the + operator one operand is String. So the other's toString() is called and added to the first. Since the 2nd operand id 'char' primitive a 'Character' Wrapper class is created automatically and its toString() is called. The toString() of Charactor returns the charactor itself. NOT the int value. So the result of the above statement is char:3

YOu should get 'Explicit cast needed to convert int to byte' compile time error for the above statement.

Regarding the next qstn, b3+=1; //compiles and runs ok,
the compound operators (anyoperand)= (+= , -= ,*= , /= , %= etc)
will convert the result automatically back to the type of the left hand side var.It is equivalent of the foll.
b3+=1; is same as b3 = (byte)(b3+1);
Simillarly
b3 += i; also will work; It is SAME as b3 = (byte)(b3+i);

regds
maha anna

While working out this example, I stumbled on this:

int i = 1;
byte b = 0;
byte b1 = 2;

b += i; // Works okay
b1 = b1 + i; // Fails to compile int cannot be converted to byte.

Whats' the difference?
( Not sure if I have to start a new thread ....)

Thanks.

- satya

OOPS, looks like both of us are online at the same time ...I gotit ...

- satya

int i =1; 
char c = '2';
i += c;
System.out.println(i); //prints 51
int i =1; 
char c = '2';
c += i;
System.out.println(c); //prints 3

Simillarly System.out.println(int i); prints the integer value to the console. Please refer to the API and do some research.
regds
maha anna