loss in precision in JAVA while using byte datatype

Question

byte b=9 ;
b=b+6 ;

gives compilation error (possible loss of precision ) why does b=9 not give error whereas b=b+9 give loss

Answer 1

During expression evaluation there is an implicit type casting to int, due to the promotion rules in java

So intead of :

byte b=1;
b=b+1;  //will give error

Write:

b=(byte)b+1;

Here are some links you can refer:

Link 1

Link 2

Link 3