calling char value

[nicky priya]

Hello everyone, i have doubt in 'char' datatype. how can we call the char method into main method. for example here in this program we are calling all types of datatype such as int, double, short, and doing addition,subtraction, multiplication and division. i want to call even the char by giving any name as value how is it possible.

public class sixteen {
void add(int x, int y){
int z=x+y;
System.out.println("Addition of x and y is=" +z);
}
void Sub(int i, int j){
short z=(short) (i-j);
System.out.println("Subtraction of x and y is=" +z);
}
void multi(long x, long y){
long z=x*y;
System.out.println("Multiplication of x and y is=" +z);
}
void div(double x, double y){
double z=x/y;
System.out.println("Division of x and y is=" +z);
}
void name(char James, char Gosling){
char z= James ;
char i= Gosling;
System.out.println("My name is" +z+ " "+i);
}

      public static void main(String[] args) {
		// TODO Auto-generated method stub
    	  sixteen arthemetic=new sixteen();
    	  arthemetic.add(42, 55);
    	  arthemetic.Sub(2, 1);
    	  arthemetic.multi(25,25);
    	  arthemetic.div(0.5, 5.56789);
    	  arthemetic.name(z);// z cannot be resolved to a variable

	}

}

[Carey Brown]

When posting code please high light your code and use the "Code" button above the edit box. I fixed it for you this time.

[Carey Brown]

On line 19 in the name() method you declare a char variable with the name 'z'. This variable is local to and only visible inside name(). Your main() method does not have a 'z' variable so you are getting this error.

In addition your name() method is expecting two char values and in main() you are only calling it with one. You'd need something like:

name( 'a', 'b' );

Also note that "James" is a String, not a char, so you can't do

name( "James", "Gosling" );

[nicky priya]

color=black] [/color]

i am sorry i'll do it from next time Thank you.[

[nicky priya]

Can I get out put as  name such as 'James Gosling'  instead of a b.

[nicky priya]

Okay i got it. Thank you very much.

[Carey Brown]

You're welcome.

[nicky priya]