help me- loss of precision error??

[j2vdk]

i want to add 2 byte variables

getting loss of precision error,,,,,

how to add them,,,,


class bytedemo{
public static void main(String ar[]){
byte a=10;
byte b=5;
byte c=a+b;
System.out.println(c);
}
}

[Eranga]

Change this as follows.

byte c= (byte) (a + b);

All variable lower that int are assign to int, unless you don't have specify.

Check the following code segment.

byte a=10;
byte b=5;
int c=a+b;
System.out.println(c);

[j2vdk]

thank you sir

[j2vdk]

Change this as follows.

byte c= (byte) (a + b);

All variable lower that int are assign to int, unless you don't have specify.

Check the following code segment.

byte a=10;
byte b=5;
int c=a+b;
System.out.println(c);

can u tell me the order, which u say lower , higher,,,,

from wer to wer conversions happen automatically,,, and in which cases we have to specify explicitly

[Eranga]

from wer to wer conversions happen automatically,,, and in which cases we have to specify explicitly

Conversion is happen automatically, if you are not specifically define.

byte and short variables are automatically convert into int, if you are not specifically define. See the example below.

    short a = 12;
    short b = 23;
    short c = a + b;

In third line you are looking a short value, as c. But actually what you have found on right side is int value. Once you added two short value, compiler assign the value to an int, looking to add in 32 bits.

Same thing is happen for byte variables as well, I'll explain in the previous code.

On the other hand, float values also assign to double.

If you want to stuck with the same type, need explicit cast.

[j2vdk]

Conversion is happen automatically, if you are not specifically define.

byte and short variables are automatically convert into int, if you are not specifically define. See the example below.

    short a = 12;
    short b = 23;
    short c = a + b;

In third line you are looking a short value, as c. But actually what you have found on right side is int value. Once you added two short value, compiler assign the value to an int, looking to add in 32 bits.

Same thing is happen for byte variables as well, I'll explain in the previous code.

On the other hand, float values also assign to double.

If you want to stuck with the same type, need explicit cast.

thank u ,,, its clear now

[Eranga]

It's pleasure to help.