solve this is my problem byte a=40,byte b=50 both add value is 90 this is with i

solve this is my problem byte a=40,byte b=50 both add value is 90 this is with in range of byte... byte range is -128to 127....
why this pgm gives error like type mismatch....

package javapgms;

public class byte1 {

public static void main(String args[])
{
byte a=40,b=50;
byte c=a+b;

System.out.println(c);
}

}
note : dont use int k...
a,b,c are in byte range... mind it..

Question Posted / sathya

2 Answers
6295 Views
I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

solve this is my problem byte a=40,byte b=50 both add value is 90 this is with in range of byte... b..

Answer / chaya k

There is a complile time error occurs while adding byte variables a and b together. Because we have to know that when we add byte numbers we get int as a resultant datatype.

Here,both a and b are in byte, then resultant datatype after addition is in int datatype but not in byte.We can't store value from int datatype to value in byte datatype so we do down-typecasting as:

byte c=(byte) a+b;

If you don't want to downcaste you should store resultant value in int datatype only as,

int c=a+b;

Is This Answer Correct ?

7 Yes

1 No

solve this is my problem byte a=40,byte b=50 both add value is 90 this is with in range of byte... b..

Answer / siva

c=(byte)(a+b);

Is This Answer Correct ?

0 Yes

0 No

Post New Answer